Solving systems of trigonometric inequalities. Solving inequalities online
Algebra project “Solving trigonometric inequalities” Completed by student of class 10 “B” Kazachkova Yulia Supervisor: mathematics teacher Kochakova N.N.
Goal To consolidate the material on the topic “Solving trigonometric inequalities” and create a reminder for students to prepare for the upcoming exam.
Objectives: Summarize the material on this topic. Systematize the information received. Consider this topic in the Unified State Exam.
Relevance The relevance of the topic I have chosen lies in the fact that tasks on the topic “Solving trigonometric inequalities” are included in the tasks of the Unified State Exam.
Trigonometric inequalities An inequality is a relation connecting two numbers or expressions through one of the signs: (greater than); ≥ (greater than or equal to). A trigonometric inequality is an inequality involving trigonometric functions.
Trigonometric inequalities The solution of inequalities containing trigonometric functions is reduced, as a rule, to the solution of the simplest inequalities of the form: sin x>a, sin x a, cos x a, tg x a,ctg x
Algorithm for solving trigonometric inequalities On an axis corresponding to a given one trigonometric function, mark the given numerical value of this function. Draw a line through the marked point intersecting the unit circle. Select the intersection points of a line and a circle, taking into account the strict or non-strict inequality sign. Select the arc of the circle on which the solutions to the inequality are located. Determine the angle values at the starting and ending points of the circular arc. Write down the solution to the inequality taking into account the periodicity of the given trigonometric function.
Formulas for solving trigonometric inequalities sinx >a; x (arcsin a + 2πn; π- arcsin a + 2πn). sinx a; x (- arccos a + 2πn; arccos a + 2πn). cosxa; x (arctg a + πn ; + πn). tgx a; x (πn ; arctan + πn). ctgx
Graphical solution of basic trigonometric inequalities sinx >a
Graphical solution of basic trigonometric inequalities sinx Graphical solution of basic trigonometric inequalities cosx >a Graphical solution of basic trigonometric inequalities cosx Graphical solution of basic trigonometric inequalities tgx >a Graphical solution of basic trigonometric inequalities tgx Graphic solution of basic trigonometric inequalities ctgx >a
Inequalities are relations of the form a › b, where a and b are expressions containing at least one variable. Inequalities can be strict - ‹, › and non-strict - ≥, ≤.
Trigonometric inequalities are expressions of the form: F(x) › a, F(x) ‹ a, F(x) ≤ a, F(x) ≥ a, in which F(x) is represented by one or more trigonometric functions.
An example of the simplest trigonometric inequality is: sin x ‹ 1/2. It is customary to solve such problems graphically; two methods have been developed for this.
Method 1 - Solving inequalities by graphing a function
To find an interval that satisfies the conditions inequality sin x ‹ 1/2, you must perform the following steps:
- On the coordinate axis, construct a sinusoid y = sin x.
- On the same axis, draw a graph of the numerical argument of the inequality, i.e., a straight line passing through the point ½ of the ordinate OY.
- Mark the intersection points of the two graphs.
- Shade the segment that is the solution to the example.
When strict signs are present in an expression, the intersection points are not solutions. Since the smallest positive period of a sinusoid is 2π, we write the answer as follows:
If the signs of the expression are not strict, then the solution interval must be enclosed in square brackets - . The answer to the problem can also be written as the following inequality: 
Method 2 - Solving trigonometric inequalities using the unit circle
Similar problems can be easily solved using a trigonometric circle. The algorithm for finding answers is very simple:
- First you need to draw a unit circle.
- Then you need to note the value of the arc function of the argument of the right side of the inequality on the arc of the circle.
- It is necessary to draw a straight line passing through the value of the arc function parallel to the abscissa axis (OX).
- After that, all that remains is to select the arc of a circle, which is the set of solutions to the trigonometric inequality.
- Write down the answer in the required form.
Let us analyze the stages of the solution using the example of the inequality sin x › 1/2. Points α and β are marked on the circle - values
The points of the arc located above α and β are the interval for solving the given inequality.
If you need to solve an example for cos, then the answer arc will be located symmetrically to the OX axis, not OY. You can consider the difference between the solution intervals for sin and cos in the diagrams below in the text.
Graphical solutions for tangent and cotangent inequalities will differ from both sine and cosine. This is due to the properties of functions.
Arctangent and arccotangent are tangents to a trigonometric circle, and the minimum positive period for both functions is π. To quickly and correctly use the second method, you need to remember on which axis the values of sin, cos, tg and ctg are plotted.
The tangent tangent runs parallel to the OY axis. If we plot the value of arctan a on the unit circle, then the second required point will be located in the diagonal quarter. Angles
They are break points for the function, since the graph tends to them, but never reaches them.
In the case of cotangent, the tangent runs parallel to the OX axis, and the function is interrupted at points π and 2π.
Complex trigonometric inequalities
If the argument of the inequality function is represented not just by a variable, but by an entire expression containing an unknown, then we are already talking about complex inequality. The process and procedure for solving it are somewhat different from the methods described above. Suppose we need to find a solution to the following inequality:
The graphical solution involves constructing an ordinary sinusoid y = sin x using arbitrarily selected values of x. Let's calculate a table with coordinates for the control points of the graph:
The result should be a beautiful curve.
To make finding a solution easier, let’s replace the complex function argument
1. If the argument is complex (different from X), then replace it with t.
2. We build in one coordinate plane tOy function graphs y=cost And y=a.
3. We find such two adjacent points of intersection of graphs, between which is located above the straight line y=a. We find the abscissas of these points.
4. Write a double inequality for the argument t, taking into account the cosine period ( t will be between the found abscissas).
5. Make a reverse substitution (return to the original argument) and express the value X from the double inequality, we write the answer in the form of a numerical interval.
Example 1.
Next, according to the algorithm, we determine those values of the argument t, at which the sinusoid is located higher straight. Let's write these values as a double inequality, taking into account the periodicity of the cosine function, and then return to the original argument X.
Example 2.
Selecting a range of values t, in which the sinusoid is above the straight line.
We write the values in the form of double inequality t, satisfying the condition. Do not forget that the smallest period of the function y=cost equals 2π. Returning to the variable X, gradually simplifying all parts of the double inequality.
We write the answer in the form of a closed numerical interval, since the inequality was not strict.
Example 3.
We will be interested in the range of values t, at which the points of the sinusoid will lie above the straight line.
Values t write it in the form of a double inequality, rewrite the same values for 2x and express X. Let's write the answer in the form of a numerical interval.
And again formula cost>a.
If cost>a, (-1≤A≤1), then - arccos a + 2πn< t < arccos a + 2πn, nєZ.
Apply formulas to solve trigonometric inequalities and you will save time on exam testing.
And now formula
, which you should use on the UNT or Unified State Examination when solving a trigonometric inequality of the form cost
If cost , (-1≤A≤1), then arccos a + 2πn< t < 2π — arccos a + 2πn, nєZ.
Apply this formula to solve the inequalities discussed in this article, and you will get the answer much faster and without any graphs!
Taking into account the periodicity of the sine function, we write a double inequality for the values of the argument t, satisfying the last inequality. Let's return to the original variable. Let us transform the resulting double inequality and express the variable X. Let's write the answer in the form of an interval.
Let's solve the second inequality:
When solving the second inequality, we had to transform the left side of this inequality using the double argument sine formula to obtain an inequality of the form: sint≥a. Next we followed the algorithm.
We solve the third inequality:
Dear graduates and applicants! Keep in mind that methods for solving trigonometric inequalities, such as the graphical method given above and, probably known to you, the method of solving using a unit trigonometric circle (trigonometric circle) are applicable only in the first stages of studying the section of trigonometry “Solving trigonometric equations and inequalities.” I think you will remember that you first solved the simplest trigonometric equations using graphs or a circle. However, now you wouldn't think of solving trigonometric equations this way. How do you solve them? That's right, according to the formulas. So trigonometric inequalities should be solved using formulas, especially during testing, when every minute is precious. So, solve the three inequalities of this lesson using the appropriate formula.
If sint>a, where -1≤ a≤1, then arcsin a + 2πn< t < π — arcsin a + 2πn, nєZ.
Learn formulas!
And finally: did you know that mathematics is definitions, rules and FORMULAS?!
Of course you do! And the most curious, having studied this article and watched the video, exclaimed: “How long and difficult! Is there a formula that allows you to solve such inequalities without any graphs or circles?” Yes, of course there is!
TO SOLVING INEQUALITIES OF THE FORM: sin (-1≤A≤1) the formula is valid:
— π — arcsin a + 2πn< t < arcsin a + 2πn, nєZ.
Apply it to the examples discussed and you will get the answer much faster!
Conclusion: LEARN FORMULAS, FRIENDS!
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1.5 Trigonometric inequalities and methods for solving them
1.5.1 Solving simple trigonometric inequalities
Most authors of modern mathematics textbooks suggest starting to consider this topic by solving the simplest trigonometric inequalities. The principle of solving the simplest trigonometric inequalities is based on the knowledge and skills of determining on a trigonometric circle the values of not only the main trigonometric angles, but also other values.
Meanwhile, the solution to inequalities of the form , , , can be carried out as follows: first we find some interval () on which this inequality is satisfied, and then write down the final answer by adding to the ends of the found interval a number that is a multiple of the period of the sine or cosine: ( 
). In this case, the value is easy to find, because or . The search for meaning is based on students’ intuition, their ability to notice the equality of arcs or segments, taking advantage of the symmetry of individual parts of the sine or cosine graph. And this is sometimes beyond the capabilities of quite a large number of students. In order to overcome the noted difficulties in textbooks in last years different approaches were used to solve the simplest trigonometric inequalities, but this did not provide any improvement in learning results.
For a number of years, we have been quite successfully using formulas for the roots of the corresponding equations to find solutions to trigonometric inequalities.
We study this topic in the following way:
1. We build graphs and y = a, assuming that .
Then we write down the equation and its solution. Giving n 0; 1; 2, we find the three roots of the compiled equation: . The values are the abscissa of three consecutive points of intersection of the graphs and y = a. It is obvious that the inequality always holds on the interval (), and the inequality always holds on the interval ().
By adding to the ends of these intervals a number that is a multiple of the period of the sine, in the first case we obtain a solution to the inequality in the form: ; and in the second case, a solution to the inequality in the form:
Only in contrast to the sine from the formula, which is a solution to the equation, for n = 0 we obtain two roots, and the third root for n = 1 in the form 
. And again, they are three consecutive abscissas of the points of intersection of the graphs and . In the interval () the inequality holds, in the interval () the inequality
Now it is not difficult to write down the solutions to the inequalities and . In the first case we get: ;
and in the second: .
Summarize. To solve the inequality or, you need to create the corresponding equation and solve it. From the resulting formula, find the roots of and , and write the answer to the inequality in the form: .
When solving inequalities , from the formula for the roots of the corresponding equation we find the roots and , and write the answer to the inequality in the form: .
This technique allows you to teach all students how to solve trigonometric inequalities, because This technique relies entirely on skills that students have a strong command of. These are the skills to solve simple problems and find the value of a variable using a formula. In addition, it becomes completely unnecessary to carefully solve a large number of exercises under the guidance of a teacher in order to demonstrate all sorts of reasoning techniques depending on the sign of the inequality, the value of the modulus of the number a and its sign. And the process of solving inequality itself becomes brief and, which is very important, uniform.
Another advantage of this method is that it allows you to easily solve inequalities even when the right side is not a table value of sine or cosine.
Let's demonstrate this with a specific example. Suppose we need to solve an inequality. Let's create the corresponding equation and solve it:
Let's find the values of and .
When n = 1 
When n = 2 
We write down the final answer to this inequality:
In the considered example of solving the simplest trigonometric inequalities, there can be only one drawback - the presence of a certain amount of formalism. But if everything is assessed only from these positions, then it will be possible to accuse the formulas of the roots of the quadratic equation, and all formulas for solving trigonometric equations, and much more, of formalism.
Although the proposed method occupies a worthy place in the formation of skills in solving trigonometric inequalities, the importance and features of other methods for solving trigonometric inequalities cannot be underestimated. These include the interval method.
Let's consider its essence.
Set edited by A.G. Mordkovich, although you shouldn’t ignore the rest of the textbooks either. § 3. Methodology for teaching the topic “Trigonometric functions” in the course of algebra and beginnings of analysis In the study of trigonometric functions at school, two main stages can be distinguished: ü Initial acquaintance with trigonometric functions...
In carrying out the research, the following tasks were solved: 1) The current textbooks of algebra and the beginnings of mathematical analysis were analyzed to identify the methods presented in them for solving irrational equations and inequalities. The analysis allows us to draw the following conclusions: ·in secondary school, insufficient attention is paid to methods for solving various irrational equations, mainly...
The simplest trigonometric inequalities of the form sin x>a are the basis for solving more complex trigonometric inequalities.
Let's consider solving the simplest trigonometric inequalities of the form sin x>a on the unit circle.
1) at 0
Using the association cosine-bun (both begin with co-, both are “round”), we remember that cosine is x, respectively, sine is y. From here we build a graph y=a - a straight line parallel to the ox axis. If the inequality is strict, the points of intersection of the unit circle and the straight line y=a are punctured, if the inequality is not strict, we paint over the points (how easy it is to remember when a point is punctured and when it is shaded, see). The greatest difficulty in solving the simplest trigonometric inequalities is caused by correctly finding the points of intersection of the unit circle and the line y=a.
The first point is easy to find - it is arcsin a. We determine the path along which we go from the first point to the second. On the line y=a sinx=a, above, above the line, sin x>a, and below, below the line, sin x
2) a=0, that is sin x>0
In this case, the first point of the interval is 0, the second is n. To both ends of the interval, taking into account the period of the sine, we add 2n.
3) for a=-1, that is sinx>-1
In this case, the first point is p/2, and to get to the second, we go around the entire circle counterclockwise. We get to the point -p/2+2p=3p/2. To take into account all intervals that are solutions to this inequality, we add 2n to both ends.
4) sinx>-a, at 0
The first point is, as usual, arcsin(-a)=-arcsina. To get to the second point, we go the upper way, that is, in the direction of increasing the angle.
This time we are moving beyond n. How long are we going to? On arcsin x. This means that the second point is n+arcsin x. Why is there no minus? Because the minus in the notation -arcsin a means clockwise movement, but we went counterclockwise. And finally, add 2pn to each end of the interval.
5) sinx>a, if a>1.
The unit circle lies entirely under the straight line y=a. There is not a single point above the straight line. So there are no solutions.
6) sinx>-a, where a>1.
In this case, the entire unit circle lies entirely above the straight line y=a. Therefore, any point satisfies the condition sinx>a. This means x is any number.
And here x is any number, since the points -n/2+2nn are included in the solution, in contrast to the strict inequality sinx>-1. There is no need to exclude anything.
The only point on the circle satisfying this condition, is p/2. Taking into account the period of the sine, the solution to this inequality is the set of points x=n/2+2n.
For example, solve the inequality sinx>-1/2:
