Methods for solving trigonometric inequalities and their systems. The simplest and complex trigonometric inequalities

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 Plotting a Function

To find an interval that satisfies the conditions of the inequality sin x ‹ 1/2, you must do the following:

On the coordinate axis build a sinusoid y = sin x.
Draw a graph on the same axis numeric argument inequality, i.e., a straight line passing through the point ½ of the y-ordinate.
Mark the intersection points of the two graphs.
Shade the segment that is the solution of the example.

When there are strong signs in an expression, the intersection points are not solutions. Since the smallest positive period sinusoid is 2π, then we write the answer as follows:

If the signs of the expression are not strict, then the interval of solutions must be enclosed in square brackets- . The answer to the problem can also be written as another inequality:

Method 2 - Solving trigonometric inequalities using the unit circle

Similar problems can be easily solved with the help of trigonometric circle. The search algorithm is very simple:

First, 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 x-axis (OX).
After that, it remains only to select the arc of a circle, which is the set of solutions to the trigonometric inequality.
Write the answer in the required form.

Let us analyze the solution steps using the inequality sin x › 1/2 as an example. Points α and β are marked on the circle – the 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 arc of answers will be located symmetrically to the OX axis, and 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.

Arc tangent and arc tangent are tangents to trigonometric circle, and the minimum positive period for both functions is π. In order to quickly and correctly use the second method, you need to remember on which axis the sin values, cos, tg and ctg.

The tangent tangent runs parallel to the OY axis. If we plot the value of arctg a on the unit circle, then the second required point will be located in the diagonal quarter. corners

They are breakpoints for the function, as the graph tends to them but never reaches them.

In the case of the cotangent, the tangent runs parallel to the OX axis, and the function is interrupted at the 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 course and order of its solution are somewhat different from the methods described above. Suppose we need to find a solution to the following inequality:

The graphical solution provides for the construction of an ordinary sinusoid y = sin x for arbitrarily chosen values of x. Let's calculate a table with coordinates for the chart's reference points:

The result should be a nice curve.

For ease of finding a solution, we replace the complex function argument

The algorithm for solving the simplest trigonometric inequalities and recognizing ways to solve trigonometric inequalities.

Teachers of the highest qualification category:

Shirko F.M. Progress village, MOBU-SOSH №6

Sankina L.S. Armavir, PEI secondary school " New way»

Does not exist universal tricks teaching disciplines of the natural-mathematical cycle. Each teacher finds his own ways of teaching acceptable only to him.

Our many years of teaching experience shows that students can more easily learn material that requires concentration and storage of a large amount of information in memory if they are taught to use algorithms in their work. initial stage learning difficult topic. Such a topic, in our opinion, is the topic of solving trigonometric inequalities.

So, before we start with students to identify techniques and methods for solving trigonometric inequalities, we work out and fix the algorithm for solving the simplest trigonometric inequalities.

Algorithm for solving the simplest trigonometric inequalities

We mark points on the corresponding axis ( for sin x- y axis, forcos x- OX axis)

We restore the perpendicular to the axis, which will intersect the circle at two points.

First on the circle we sign the point that belongs to the interval of the range of values of the arc function by definition.

Starting from the signed point, we shade the arc of a circle corresponding to the shaded part of the axis.

We turn Special attention in the direction of bypass. If the traversal is clockwise (i.e. there is a transition through 0), then the second point on the circle will be negative, if counterclockwise - positive.

We write the answer as an interval, taking into account the periodicity of the function.

Let's consider the operation of the algorithm with examples.

1) sin ≥ 1/2;

Decision:

Draw a unit circle.;

We mark a point ½ on the y-axis.

Restore the perpendicular to the axis,

which intersects the circle at two points.

By the definition of the arcsine, we mark first

point π/6.

We shade the part of the axis that corresponds to

given inequality, above the point ½.

We shade the arc of a circle corresponding to the shaded part of the axis.

The bypass is made counterclockwise, we got the point 5π/6.

We write the answer as an interval, taking into account the periodicity of the function;

Answer:x;[π/6 + 2π n, 5π/6 + 2π n], n Z.

The simplest inequality is solved using the same algorithm if there is no tabular value in the answer record.

Students, in the first lessons, solving inequalities at the blackboard, pronounce each step of the algorithm aloud.

2) 5 cos x – 1 ≥ 0;

R Solution:at

5 cos x – 1 ≥ 0;

cos x ≥ 1/5;

Draw a unit circle.

We mark on the OX axis a point with the coordinate 1/5.

We restore the perpendicular to the axis, which

intersects the circle at two points.

First on the circle we sign the point that belongs to the interval of the range of values of the arccosine by definition (0; π).

We shade the part of the axis that corresponds to this inequality.

Starting from signed point arccos 1/5, shade the arc of a circle corresponding to the shaded part of the axis.

The bypass is made clockwise (i.e. there is a transition through 0), which means that the second point on the circle will be negative - arccos 1/5.

We write the answer as an interval, taking into account the periodicity of the function, from a smaller value to a larger one.

Answer: x  [-arccos 1/5 + 2π n, arccos 1/5 + 2π n], n Z.

Improving the ability to solve trigonometric inequalities is facilitated by the questions: “How will we solve a group of inequalities?”; “How does one inequality differ from another?”; “How is one inequality similar to another?”; How would the answer change if a strict inequality were given? How would the answer change if there was a sign instead of the sign ""

The task of analyzing the list of inequalities from the standpoint of ways to solve them allows you to work out their recognition.

Students are given inequalities to solve in class.

Question: Highlight the inequalities that need to be applied equivalent transformations when reducing the trigonometric inequality to the simplest?

Answer 1, 3, 5.

Question: What are the inequalities in which it is required to consider a complex argument as a simple one?

Answer: 1, 2, 3, 5, 6.

Question: Name the inequalities where you can apply trigonometric formulas?

Answer: 2, 3, 6.

Question: What are the inequalities where you can apply the method of introducing a new variable?

Answer: 6.

The task of analyzing the list of inequalities from the standpoint of ways to solve them allows you to work out their recognition. When developing skills, it is important to single out the stages of its implementation and formulate them in general view, which is presented in the algorithm for solving the simplest trigonometric inequalities.