Lesson model on the topic:
"Solving trigonometric equations and inequalities"
as part of the implementation of the regional component in mathematics
for 10th grade students.
Pomykalova
Elena Viktorovna
mathematic teacher
Municipal educational institution secondary school of the village of Voskhod
Balashovsky district
Saratov region
The purpose of the lesson.
1. Summarize theoretical knowledge on the topic: “Solving trigonometric equations and inequalities”, repeat the basic methods for solving trigonometric equations and inequalities.
2. Develop the qualities of thinking: flexibility, focus, rationality. Organize students’ work on the specified topic at a level corresponding to the level of knowledge already formed.
3. Cultivate accuracy of notes, culture of speech, and independence.
Lesson type: a lesson in generalizing and systematizing the knowledge acquired while studying this topic.
Teaching methods: system generalization, test checking the level of knowledge, solving generalization problems.
Forms of lesson organization: frontal, individual.
Equipment: computer , multimedia projector, answer sheets, task cards, table of formulas for roots of trigonometric equations.
During the classes.
I . Start of the lesson
The teacher informs students about the topic of the lesson, the purpose, and draws students' attention to the handouts.
II . Monitoring student knowledge
1) Oral work (The task is projected onto the screen)
Calculate:
A) ;
b) ;
V) ;
G) ;
d) ;
e) .
2) Frontal survey of students.
What equations are called trigonometric?
What types of trigonometric equations do you know?
What equations are called the simplest trigonometric equations?
What equations are called homogeneous?
What equations are called quadratic?
What equations are called inhomogeneous?
What methods of solving trigonometric equations do you know?
After students answer, some ways to solve trigonometric equations are projected on the screen.
Introducing a new variable:
№1 . 2sin²x – 5sinx + 2 = 0.№2. tg + 3ctg = 4.
Let sinx = t, |t|≤1, Let tg = z,
We have: 2 t² – 5 t + 2 = 0. We have: z + = 4.
2. Factorization :
2 sinxcos 5 x – cos 5 x = 0;
cos5x (2sinx – 1) = 0.
We have : cos5x = 0,
2sinx – 1 = 0; ...
3. Homogeneous trigonometric equations:
I degrees II degrees
a sinx + b cosx = 0, (a,b ≠ 0). a sin²x + b sinx cosx + c cos²x = 0.
Divide by cosx≠ 0. 1) if a ≠ 0, divide bycos² x ≠ 0
We have : a tgx + b = 0; ...we have : a tg²x + b tgx + c = 0.
2) if a = 0, then
we have: bsinxcosx + ccos² x =0;…
4. Inhomogeneous trigonometric equations:
Equations of the form: asinx + bcosx = c
4 sinx + 3 cosx = 5.
(Show two ways)
1) use of universal substitution:
sinx = (2 tgx/2) / (1 + tg 2 x/2);
cosx = (1– tg 2 x/2) / (1 + tg 2 x/2);
2) introducing an auxiliary argument:
4 sinx + 3 cosx = 5
Divide both sides by 5:
4/5 sinx + 3/5 cosx = 1
Since (4/5) 2 + (3/5) 2 = 1, then let 4/5 = sinφ; 3/5= cosφ, where 0< φ < π /2, then
sinφsinx + cosφcosx = 1
cos(x – φ ) = 1
x – φ = 2 πn, n € Z
x = 2 πn + φ , n € Z
φ = arccos 3/5 means x = arcos 3/5 +2 πn, n € Z
Answer: arccos 3/5 + 2 πn, n € Z
3) Solving equations using formulas for reducing the degree.
4) Application of double and triple argument formulas.
a) 2sin4xcos2x = 4cos 3 2x – 3cos2x
cos6x +cos2x = cos6x
III . Executing a test task
The teacher asks students to apply the theoretical facts just formulated to solve equations.
The task is carried out in the form of a test. Students fill out the answer form located on their desks.
The task is projected onto the screen.
Suggest a way to solve this trigonometric equation:
1) reduction to square;
2) reduction to homogeneity;
3) factorization;
4) reduction in degree;
5) converting the sum of trigonometric functions into a product.
Answer form.
Option I
The equation
Solutions
3 sin²x + cos²x = 1 - sinx cosx
4 co s²x- cosx– 1 = 0
2 sin² x / 2 +cosx=1
cosx + cos3x = 0
2 sinx cos5x – cos5x = 0
Option II
The equation
Solutions
2sinxcosx – sinx = 0
3 cos²x - cos2x = 1
6 sin²x + 4 sinx cosx = 1
4 sin²x + 11 sin²x = 3
sin3x = sin17x
Answers:
Option I Option II
IV . Repeating formulas to solve equations
Formulas for roots of trigonometric equations.
Are common
Private
The equation
Root formula
The equation
Root formula
1. sinx = a, |a|≤1
x = (-1) n arcsin a + πk,
kє Z
1. sinx = 0
x = πk, kє Z
2. cosx = a, |a|≤1
x = ±arccos a + 2πk,
kє Z
2. sinx = 1
x = + 2πk, k є Z
3. tg x = a
x = arctan a + πk, kє Z
3. sinx = –1
x = – + 2πk, k є Z
4.ctg x = a
x = arcctg a + πk,kє Z
4. cosx = 0
x = + πk, k є Z
5. cosx = 1
x = 2πk, k є Z
6. cosx = –1
x = π + 2πk, k є Z
Oral work on solving simple trigonometric equations
The teacher asks students to apply the theoretical facts just formulated to solve equations. A simulator for oral work on the topic: “Trigonometric equations” is projected onto the screen.
Solve equations.
sinx = 0
cosx = 1
tan x = 0
ctg x = 1
sin x = - 1 / 2
sin x = 1
cos x = 1 / 2
sin x = - √3 / 2
cos x = √2 / 2
sin x = √2 / 2
cos x = √3 / 2
tan x = √3
sin x = 1 / 2
sin x = -1
cos x = - 1 / 2
sin x = √3 / 2
tan x = -√3
ctg x = √3 / 3
tan x = - √3 / 3
cot x = -√3
cos x – 1 =0
2 sin x – 1 =0
2ctg x + √3 = 0
V . Solving examples.
Cards with tasks are distributed to each desk, one is on the teacher’s desk for students coming to the board.
№1. Find the arithmetic mean of all the roots of the equation , satisfying the condition ;
Solution.
Let's find the arithmetic mean of all roots of a given equation from the interval .
.
Answer: a) .
№2 . Solve the inequality .
Solution.
,
,
.
Answer: 
№3. Solve the equation .
(Jointly determine a method for solving the problem)
Solution.
Let us estimate the right and left sides of the last equality.
Therefore, equality holds if and only if the system holds
Answer: 0.5
VI . Independent work
The teacher gives tasks for independent work. Cards are prepared according to difficulty levels.
More prepared students can be given cards with tasks of an increased level of complexity.
The teacher gave the students of the 2nd group cards with tasks of a basic level of complexity.
For students of the 3rd group, the teacher compiled cards with tasks of a basic level of complexity, but these are, as a rule, students with poor mathematical preparation; they can complete tasks under the supervision of the teacher.
Along with the assignments, students receive forms to complete the assignments.
1 group
Option #1 (1)
1. Solve the equation
2. Solve the equation .
Option #2 (1)
1. Solve the equation .
2. Solve the equation .
2nd group
Option #1 (2)
1. Solve the equation .
2. Solve the equation .
To view the presentation with pictures, design and slides, download its file and open it in PowerPoint on your computer.
Text content of presentation slides: Solving trigonometric inequalities by the method of intervals 10 A class Teacher: Uskova N.N. MBOU Lyceum No. 60 Lesson objectives: Educational: expanding and deepening knowledge on the topic “Method of intervals”; acquiring practical skills in completing tasks using the interval method; increasing the level of mathematical training of schoolchildren; Developmental: development of research skills; Educational: formation of observation, independence, ability to interact with other people, nurturing a culture of thinking, a culture of speech, interest in the academic subject. Lesson progress Checking homework. Independent work. Explanation of new material on the topic “Solving trigonometric inequalities by the interval method”: solution algorithm; examples of inequalities. Lesson summary. Homework. Checking homework Solve inequalities: Independent work Additionally: 1) 2) Checking homework Solve inequalities: a) Solution. Answer: b) Solution. Answer: c) Solution. Answer: d) Solution. Answer: . Solve inequality Solution. Answer: Example 1. Solve inequality using the interval method Solution. 1) 2) Zeros of the function: 3) Signs of the function on intervals: + - + - + 4) Since the inequality is not strict, the roots are included 5) Solution: Answer: Example 2. Solve the inequality: Solution. Answer: Method I: Method II: Answer: Solving trigonometric inequalities using the interval method Algorithm: Using trigonometric formulas, factorize. Find discontinuity points and zeros of the function, place them on the circle. Take any point x0 (but not previously found) and find out the sign works. If the product is positive, then place a “+” behind the unit circle on the ray corresponding to the angle. Otherwise, put a “-” sign inside the circle. If a point occurs an even number of times, we call it a point of even multiplicity, if an odd number of times, we call it a point of odd multiplicity. Draw arcs as follows: start from point x0, if the next point is of odd multiplicity, then the arc intersects the circle at this point, but if the point is of even multiplicity, then it does not. Arcs beyond the circle are positive intervals; inside the circle there are negative spaces. Solution of examples 1) 2) 3) 4) 5) Example 1. Solution. Points of the first series: Points of the second series: - - - + + + Answer: Example 2. Solution. Points of the first series: Points of the second series: Points of the third series: Points of the fourth series: Points of even multiplicity: + + + + - - - - Answer: Example 3. Solution. Total: Points of the first series: Points of the second series: Points of the third series: + + + + + + - - - - - - - - Answer. Points of even multiplicity: Example 4. Solution. + + + + - - - - Answer. Example 5. Solution. 1) 2) Zeros of the function: 3) + - - + - there are no zeros So, at Answer: Graphically: Homework: Solve trigonometric inequalities by the method of intervals: a) b) c) d) e) f) g) Additional tasks:
Attached files
The topic “Trigonometric inequalities” is objectively difficult for 10th grade students to perceive and comprehend. Therefore, it is very important to consistently, from simple to complex, develop an understanding of the algorithm and develop a stable skill in solving trigonometric inequalities.
The article presents an algorithm for solving the simplest trigonometric inequalities and provides a summary of a lesson in which more complex types of trigonometric inequalities are mastered.
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Shchalpegina I.V.
The topic “Trigonometric inequalities” is objectively difficult for 10th grade students to perceive and comprehend. Therefore, it is very important to consistently, from simple to complex, develop an understanding of the algorithm and develop a stable skill in solving trigonometric inequalities.
Success in mastering this topic depends on knowledge of the basic definitions and properties of trigonometric and inverse trigonometric functions, knowledge of trigonometric formulas, the ability to solve integer and fractional rational inequalities, and the main types of trigonometric equations.
Particular emphasis should be placed on the method of teaching solutions protozoa trigonometric inequalities, because any trigonometric inequality reduces to solving the simplest inequalities.
It is preferable to introduce the primary idea of solving simple trigonometric inequalities using graphs of sine, cosine, tangent and cotangent. And only then learn to solve trigonometric inequalities on a circle.
I will dwell on the main stages of reasoning when solving the simplest trigonometric inequalities.
- We find points on the circle whose sine (cosine) is equal to the given number.
- In the case of a strict inequality, we mark these points on the circle as punctured; in the case of a non-strict inequality, we mark them as shaded.
- The point lying onthe main interval of monotonysine (cosine) functions, called P t1, another point - P t2.
- We mark along the sine (cosine) axis the interval that satisfies this inequality.
- We select an arc on the circle corresponding to this interval.
- We determine the direction of movement along the arc (from point P t1 to point P t2 along an arc ), we draw an arrow in the direction of movement, above which we write a “+” or “-” sign, depending on the direction of movement. (This stage is important for monitoring the found angles. Students can illustrate the common mistake of finding the boundaries of an interval using the example of solving the inequality on schedule sine or cosine and around the circumference).
- Finding the coordinates of points P t1 (as arcsine or arccosine of a given number) and Р t2 those. boundaries of the interval, we control the correctness of finding the angles by comparing t 1 and t 2.
- We write the answer in the form of a double inequality (or gap) from the smaller angle to the larger one.
The reasoning for solving inequalities with tangent and cotangent is similar.
The drawing and recording of the solution, which should be reflected in the students’ notebooks, are given in the proposed outline.
Lesson summary on the topic: “Solving trigonometric inequalities.”
Lesson Objective – continue studying the solution of trigonometric inequalities containing the functions sine and cosine, move from the simplest inequalities to more complex ones.
Lesson objectives:
- consolidation of knowledge of trigonometric formulas, tabular values of trigonometric functions, formulas for the roots of trigonometric equations;
- developing the skill of solving simple trigonometric inequalities;
- mastering techniques for solving more complex trigonometric inequalities;
- development of logical thinking, semantic memory, independent work skills, self-testing;
- fostering accuracy and clarity in the formulation of solutions, interest in the subject, respect for classmates.
- formation of educational, cognitive, information, and communication competencies.
Equipment: graph projector, handout cards with ready-made drawings of trigonometric circles, portable board, cards with homework.
Form organization of training - lesson. Methods teaching used in the lesson - verbal, visual, reproductive, problem-search, individual and frontal questioning, oral and written self-control, independent work.
N p/p | Lesson stages. | |||||||||
Organizing a class for work. | ||||||||||
Checking homework. | (Collecting notebooks with homework) |
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Statement of the purpose of the lesson. | Today in the lesson we will repeat the solution of the simplest trigonometric inequalities and consider more complex cases. |
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Oral work. | (Tasks and answers are written on an overhead projector tape, I open the answers as I solve them)
sinx = -, 2sinx =, sin2x = , sin(x -) = 0, cosx = , cosx = -, cos2x = 1, tgx = -1.
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Repetition. | Let us recall the algorithm for solving the simplest trigonometric inequalities. (On the board there are blanks of two circles. I call two students one at a time to solve inequalities. The student explains in detail the solution algorithm. The class works together with those answering at the board on pre-prepared cards with the image of a circle).
How does the solution of strict inequality affect the answer? (3) and 4) two students solve inequalities on an overhead projector tape, the class solves them independently on cards).
Swap options, take a pen of a different color, check your friend’s work. (Self-test from an overhead projector tape. The student completing the task comments on the solution. After returning the work, reflection). How does the solution to the inequality change when the argument x is replaced by 2x, by? (Assessing student work). |
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New material. | Let's move on to more complex trigonometric inequalities, the solution of which will be reduced to solving the simplest trigonometric inequalities. Let's look at examples. (Solving inequalities on the board under the guidance of the teacher). No. 1. cos 2 2x – 2cos2x ≥ 0. (Let us recall the technique of solving trigonometric equations by placing the common factor out of brackets). cos2x(cos2x – 2) ≥ 0. Replacement: cos2x = t, ≤ 1; t(t – 2) ≥ 0;The second inequality does not satisfy the condition ≤ 1. cos2x ≤ 0. (Solve the inequality yourself. Check the answer). Answer: + n x + n, n Z. No. 2. 6sin 2 x – 5sinx + 1 ≥ 0. (Remember the technique of solving trigonometric equations by changing a variable. The student solves it at the board with comments). Replacement sinx = t, ≤ 1. 6t 2 – 5t +1 ≥ 0.6(t -)(t -), Answer: + 2 n ≤ x ≤ + 2 n, - -arcsin+ 2 k ≤ x ≤ arcsin+ 2 k, n, k Z. No. 3. sinx + cos2x 1. (We discuss solution options. We recall the formula for the cosine of a double angle. The class decides independently, one student - on an individual board, followed by verification). sinx + cos2x - 1 0, sinx – 2sin 2 x 0, sinx(1 - 2 sinx) 0,
Analyze situations when the answer to solving a quadratic inequality is written in the form of a set of two inequalities, and when – in the form of a system. The following diagram is useful: No. 4. coscosx - sinsinx -. (Discussion. One student is called to the board for each step of the solution, the stages are commented on. The teacher checks the recording with the students working on the spot). cos(x +) -, cost -.
No. 5. Define everything A , for each of which the inequality 4sinx + 3cosx ≤ a has at least one solution. (Remember the algorithm for solving a trigonometric equation with a normalizing factor. The solution is written on an overhead projector tape. I open it step by step as I reason. Differentiated work). 4sinx + 3cosx ≤ a , M = = 5. Divide both sides of the inequality by 5: sinx + cosx ≤ . Because () 2 + () 2 = 1, then there is an angle α such that cosα = and sinα = . Let's rewrite the previous inequality in the form: sin(x + α) ≤ . The last inequality, and therefore the original inequality, has at least one solution for each and such that ≥ -1, that is, for every a ≥ -5. Answer: a ≥ -5. |
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Homework. | (I hand out cards with homework written down. I comment on the solution to each inequality).
Review trigonometric addition formulas and prepare for independent work. |
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Summing up, reflection. | Name methods for solving trigonometric inequalities. How is knowledge of an algorithm for solving simple trigonometric inequalities used in solving more complex inequalities? Which inequalities caused the most difficulty? (I evaluate students’ work in class). |
Independent work
based on the results of mastering the material.
Option 1. Solve inequalities 1 – 3:
| Option 2. Solve inequalities 1 – 3:
|
LESSON TOPIC: Solving simple trigonometric inequalities
The purpose of the lesson: show an algorithm for solving trigonometric inequalities using the unit circle.
Lesson Objectives:
Educational – ensure repetition and systematization of the topic material; create conditions for monitoring the acquisition of knowledge and skills;
Developmental - to promote the formation of skills to apply techniques: comparison, generalization, identification of the main thing, transfer of knowledge to a new situation, development of mathematical horizons, thinking and speech, attention and memory;
Educational – to promote interest in mathematics and its applications, activity, mobility, communication skills, and general culture.
Students' knowledge and skills:
- know the algorithm for solving trigonometric inequalities;
Be able to solve simple trigonometric inequalities.
Equipment: interactive whiteboard, lesson presentation, cards with independent work tasks.
DURING THE CLASSES:
1. Organizational moment(1 min)
I propose the words of Sukhomlinsky as the motto of the lesson: “Today we are learning together: me, your teacher and you are my students. But in the future the student must surpass the teacher, otherwise there will be no progress in science.”
2. Warm up. Dictation “True - False”
3. Repetition
For each option - task on the slide, continue each entry. Running time 3 min.
Let's cross-check this work of ours using the answer table on the board.
Evaluation criterion:“5” - all 9 “+”, “4” - 8 “+”, “3” - 6-7 “+”
4. Updating students’ knowledge(8 min)
Today in class we must learn the concept of trigonometric inequalities and master the skills of solving such inequalities.
– Let’s first remember what a unit circle is, a radian measure of an angle, and how the angle of rotation of a point on a unit circle is related to the radian measure of an angle. (working with presentation)
Unit circle is a circle with radius 1 and center at the origin.
The angle formed by the positive direction of the axis OX and the ray OA is called the rotation angle. It's important to remember where the 0 corners are; 90; 180; 270; 360.
If A is moved counterclockwise, positive angles are obtained.
If A is moved clockwise, negative angles are obtained.
сos t is the abscissa of a point on the unit circle, sin t is the ordinate of a point on the unit circle, t is the angle of rotation with coordinates (1;0).
5 . Explanation of new material (17 min.)
Today we will get acquainted with the simplest trigonometric inequalities.
Definition.
The simplest trigonometric inequalities are inequalities of the form:
The guys will tell us how to solve such inequalities (presentation of projects by students with examples). Students write down definitions and examples in their notebooks.
During the presentation, students explain the solution to the inequality, and the teacher completes the drawings on the board.
An algorithm for solving simple trigonometric inequalities is given after the students' presentation. Students see all stages of solving an inequality on the screen. This promotes visual memorization of the algorithm for solving a given problem.
Algorithm for solving trigonometric inequalities using the unit circle:
1. On the axis corresponding to a given trigonometric function, mark the given numerical value of this function.
2. Draw a line through the marked point intersecting the unit circle.
3. Select the points of intersection of the line and the circle, taking into account the strict or non-strict inequality sign.
4. Select the arc of the circle on which the solutions to the inequality are located.
5. Determine the values of the angles at the starting and ending points of the circular arc.
6. Write down the solution to the inequality taking into account the periodicity of the given trigonometric function.
To solve inequalities with tangent and cotangent, the concept of a line of tangents and cotangents is useful. These are the lines x = 1 and y = 1, respectively, tangent to the trigonometric circle.
6. Practical part(12 min)
To practice and consolidate theoretical knowledge, we will complete small tasks. Each student receives task cards. Having solved the inequalities, you need to choose an answer and write down its number.
7. Reflection on activities in the lesson
-What was our goal?
- Name the topic of the lesson
- We managed to use a well-known algorithm
- Analyze your work in class.
8. Homework(2 minutes)
Solve the inequality:
9. Lesson summary(2 minutes)
I propose to end the lesson with the words of Y.A. Komensky: “Consider unhappy that day or that hour in which you have not learned anything new and have not added anything to your education.”
During the practical lesson we will repeat main types of tasks from the topic “Trigonometry”, we will further analyze tasks of increased complexity and consider examples of solving various trigonometric inequalities and their systems.
This lesson will help you prepare for one of the types of tasks B5, B7, C1 And C3.
Preparation for the Unified State Exam in mathematics
Experiment
Lesson 11. Consolidation of the material covered. Trigonometric inequalities. Solving various problems of increased complexity
Practice
Lesson summary
Trigonometry review
Let's start by reviewing the main types of tasks that we covered in the topic "Trigonometry" and solve several non-standard problems.
Task No. 1. Convert angles to radians and degrees: a) ; b) .
a) Let’s use the formula for converting degrees to radians
Let's substitute the specified value into it.
b) Apply the formula for converting radians to degrees
Let's perform the substitution 
.
Answer. A) ; b) .
Task No. 2. Calculate: a) ; b) .
a) Since the angle goes far beyond the table, we will reduce it by subtracting the sine period. Since the angle is indicated in radians, we will consider the period as .
b) In this case the situation is similar. Since the angle is indicated in degrees, we will consider the period of the tangent as .
The resulting angle, although smaller than the period, is larger, which means that it no longer refers to the main, but to the extended part of the table. In order not to once again train your memory by memorizing the extended table of trigofunction values, let’s subtract the tangent period again:
We took advantage of the oddness of the tangent function.
Answer. a) 1; b) .
Task No. 3. Calculate 
, If .
Let us reduce the entire expression to tangents by dividing the numerator and denominator of the fraction by . At the same time, we can not be afraid that, since in this case the tangent value would not exist.
Task No. 4. Simplify the expression.
The specified expressions are converted using reduction formulas. They are just unusually written using degrees. The first expression generally represents a number. Let's simplify all the trigofunctions one by one:
Because , the function changes to a cofunction, i.e., to a cotangent, and the angle falls into the second quarter, in which the original tangent has a negative sign.
For the same reasons as in the previous expression, the function changes to a cofunction, i.e., to a cotangent, and the angle falls into the first quarter, in which the original tangent has a positive sign.
Let's substitute everything into a simplified expression:
Problem #5. Simplify the expression.
Let us write the tangent of the double angle using the appropriate formula and simplify the expression:
The last identity is one of the universal replacement formulas for the cosine.
Problem #6. Calculate.
The main thing is not to make the standard mistake of not giving the answer that the expression is equal to . You cannot use the basic property of the arctangent as long as there is a factor in the form of two next to it. To get rid of it, we will write the expression according to the formula for the tangent of a double angle, while treating , as an ordinary argument.
Now we can apply the basic property of the arctangent; remember that there are no restrictions on its numerical result.
Problem No. 7. Solve the equation.
When solving a fractional equation that is equal to zero, it is always indicated that the numerator is equal to zero, but the denominator is not, since you cannot divide by zero.
The first equation is a special case of the simplest equation that can be solved using a trigonometric circle. Remember this solution yourself. The second inequality is solved as the simplest equation using the general formula for the roots of the tangent, but only with the sign not equal.
As we see, one family of roots excludes another family of exactly the same type of roots that do not satisfy the equation. That is, there are no roots.
Answer. There are no roots.
Problem No. 8. Solve the equation.
Let's immediately note that we can take out the common factor and let's do it:
The equation has been reduced to one of the standard forms, where the product of several factors equals zero. We already know that in this case, either one of them is equal to zero, or the other, or the third. Let's write this in the form of a set of equations:
The first two equations are special cases of the simplest ones; we have already encountered similar equations many times, so we will immediately indicate their solutions. We reduce the third equation to one function using the double angle sine formula.
Let's solve the last equation separately:
This equation has no roots, because the sine value cannot go beyond 
.
Thus, the solution is only the first two families of roots; they can be combined into one, which is easy to show on the trigonometric circle:
 |
This is a family of all halves, i.e.
Trigonometric inequalities
Let's move on to solving trigonometric inequalities. First, we will analyze the approach to solving the example without using formulas for general solutions, but using the trigonometric circle.
Problem No. 9. Solve inequality.
Let us draw an auxiliary line on the trigonometric circle corresponding to a sine value equal to , and show the range of angles that satisfy the inequality.
 |
It is very important to understand exactly how to indicate the resulting interval of angles, i.e. what is its beginning and what is its end. The beginning of the interval will be the angle corresponding to the point that we will enter at the very beginning of the interval if we move counterclockwise. In our case, this is the point that is on the left, because moving counterclockwise and passing the right point, on the contrary, we leave the required range of angles. The right point will therefore correspond to the end of the gap.
Now we need to understand the angles of the beginning and end of our interval of solutions to the inequality. A typical mistake is to immediately indicate that the right point corresponds to the angle, the left one and give the answer. This is not true! Please note that we have just indicated the interval corresponding to the upper part of the circle, although we are interested in the lower part, in other words, we have mixed up the beginning and end of the solution interval we need.
In order for the interval to start from the corner of the right point and end with the corner of the left point, it is necessary that the first specified angle be less than the second. To do this, we will have to measure the angle of the right point in the negative direction of reference, i.e. clockwise and it will be equal to . Then, starting to move from it in a positive clockwise direction, we will get to the right point after the left point and get the angle value for it. Now the beginning of the interval of angles is less than the end, and we can write the interval of solutions without taking into account the period:
Considering that such intervals will be repeated an infinite number of times after any integer number of rotations, we obtain a general solution taking into account the sine period:
We put parentheses because the inequality is strict, and we pick out the points on the circle that correspond to the ends of the interval.
Compare the answer you receive with the formula for the general solution that we gave in the lecture.
Answer. 
.
This method is good for understanding where the formulas for general solutions of the simplest trigon inequalities come from. In addition, it is useful for those who are too lazy to learn all these cumbersome formulas. However, the method itself is also not easy; choose which approach to the solution is most convenient for you.
To solve trigonometric inequalities, you can also use graphs of functions on which an auxiliary line is constructed, similar to the method shown using a unit circle. If you are interested, try to figure out this approach to the solution yourself. In what follows we will use general formulas to solve simple trigonometric inequalities.
Problem No. 10. Solve inequality.
Let us use the formula for the general solution, taking into account the fact that the inequality is not strict:
In our case we get:
Answer. 
Problem No. 11. Solve inequality.
Let us use the general solution formula for the corresponding strictly inequality:
Answer. 
.
Problem No. 12. Solve inequalities: a) ; b) .
In these inequalities, there is no need to rush to use formulas for general solutions or the trigonometric circle; it is enough to simply remember the range of values of sine and cosine.
a) Since 
, then the inequality does not make sense. Therefore, there are no solutions.
b) Since similarly, the sine of any argument always satisfies the inequality specified in the condition. Therefore, all real values of the argument satisfy the inequality.
Answer. a) there are no solutions; b) .
Problem 13. Solve inequality 
.
This simplest inequality with a complex argument is solved similarly to a similar equation. First, we find a solution for the entire argument indicated in brackets, and then we transform it to the form “”, working with both ends of the interval, as with the right side of the equation.
