# Basic Trigonometric Inequalities

Unknown variable (angle): x
Set of integers:
Integer: n
Set of real numbers:
Real number: a

Trigonometric functions: sin x, cos x, tan x, cot x
Inverse trigonometric functions: arcsin a, arccos a, arctan a, arccot a

An inequality involving trigonometric functions of an unknown angle is called a trigonometric inequality.

The following $$16$$ inequalities refer to basic trigonometric inequalities:

$\sin x \gt a,\; \sin x \ge a,\; \sin x \lt a,\; \sin x \le a,$
$\cos x \gt a,\; \cos x \ge a,\; \cos x \lt a,\; \cos x \le a,$
$\tan x \gt a,\; \tan x \ge a,\; \tan x \lt a,\; \tan x \le a,$
$\cot x \gt a,\; \cot x \ge a,\; \cot x \lt a,\; \cot x \le a.$

Here $$x$$ is an unknown variable, $$a$$ can be any real number.

## Inequalities of the form $$\sin x \gt a,$$ $$\sin x \ge a,$$ $$\sin x \lt a,$$ $$\sin x \le a$$

### Inequality $$\sin x \gt a$$

If $$\left| a \right| \ge 1$$, the inequality $$\sin x \gt a$$ has no solutions: $$x \in \varnothing.$$

If $$a \lt -1$$, the solution of the inequality $$\sin x \gt a$$ is any real number: $$x \in \mathbb{R}.$$

For $$-1 \le a \lt 1$$, the solution of the inequality $$\sin x \gt a$$ is expressed in the form

$\arcsin a + 2\pi n \lt x \lt \pi - \arcsin a + 2\pi n, \;n \in \mathbb{Z}.$

### Inequality $$\sin x \ge a$$

If $$a \gt 1$$, the inequality $$\sin x \ge a$$ has no solutions: $$x \in \varnothing.$$

If $$a \le -1$$, the solution of the inequality $$\sin x \ge a$$ is any real number: $$x \in \mathbb{R}.$$

Case $$a = 1:$$

$x = \pi/2 +2\pi n,\; n \in \mathbb{Z}.$

For $$-1 \lt a \lt 1$$, the solution of the non-strict inequality $$\sin x \ge a$$ includes the boundary angles and has the form

$\arcsin a + 2\pi n \le x \le \pi - \arcsin a + 2\pi n,\;n \in \mathbb{Z}.$

### Inequality $$\sin x \lt a$$

If $$a \gt 1$$, the solution of the inequality $$\sin x \lt a$$ is any real number: $$x \in \mathbb{R}.$$

If $$a \le -1$$, the inequality $$\sin x \lt a$$ has no solutions: $$x \in \varnothing.$$

For $$-1 \lt a \le 1$$, the solution of the inequality $$\sin x \lt a$$ lies in the interval

$-\pi - \arcsin a + 2\pi n \lt x \lt \arcsin a + 2\pi n,\;n \in \mathbb{Z}.$

### Inequality $$\sin x \le a$$

If $$a \ge 1$$, the solution of the inequality $$\sin x \le a$$ is any real number: $$x \in \mathbb{R}.$$

If $$a \lt -1$$, the inequality $$\sin x \le a$$ has no solutions: $$x \in \varnothing.$$

Case $$a = -1:$$

$x = -\pi/2 + 2\pi n,\;n \in \mathbb{Z}.$

For $$-1 \lt a \lt 1$$, the solution of the non-strict inequality $$\sin x \le a$$ is in the interval

$-\pi - \arcsin a + 2\pi n \le x \le \arcsin a + 2\pi n,\;n \in \mathbb{Z}.$

## Inequalities of the form $$\cos x \gt a,$$ $$\cos x \ge a,$$ $$\cos x \lt a,$$ $$\cos x \le a$$

### Inequality $$\cos x \gt a$$

If $$a \ge 1$$, the inequality $$\cos x \gt a$$ has no solutions: $$x \in \varnothing.$$

If $$a \lt -1$$, the solution of the inequality $$\cos x \gt a$$ is any real number: $$x \in \mathbb{R}.$$

For $$-1 \le a \lt 1$$, the solution of the inequality $$\cos x \gt a$$ has the form

$-\arccos a + 2\pi n \lt x \lt \arccos a + 2\pi n,\;n \in \mathbb{Z}.$

### Inequality $$\cos x \ge a$$

If $$a \gt 1$$, the inequality $$\cos x \ge a$$ has no solutions: $$x \in \varnothing.$$

If $$a \le -1$$, the solution of the inequality $$\cos x \ge a$$ is any real number: $$x \in \mathbb{R}.$$

Case $$a = 1:$$

$x = 2\pi n,\;n \in \mathbb{Z}.$

For $$-1 \lt a \lt 1$$, the solution of the non-strict inequality $$\cos x \ge a$$ is expressed by the formula

$-\arccos a + 2\pi n \le x \le \arccos a + 2\pi n,\;n \in \mathbb{Z}.$

### Inequality $$\cos x \lt a$$

If $$a \gt 1$$, the inequality $$\cos x \lt a$$ is true for any real value of $$x$$: $$x \in \mathbb{R}.$$

If $$a \le -1$$, the inequality $$\cos x \lt a$$ has no solutions: $$x \in \varnothing.$$

For $$-1 \lt a \le 1$$, the solution of the inequality $$\cos x \lt a$$ is written in the form

$\arccos a + 2\pi n \lt x \lt 2\pi - \arccos a + 2\pi n,\;n \in \mathbb{Z}.$

### Inequality $$\cos x \le a$$

If $$a \ge 1$$, the solution of the inequality $$\cos x \le a$$ is any real number: $$x \in \mathbb{R}.$$

If $$a \lt -1$$, the inequality $$\cos x \le a$$ has no solutions: $$x \in \varnothing.$$

Case $$a = -1:$$

$x = \pi + 2\pi n,\;n \in \mathbb{Z}.$

For $$-1 \lt a \lt 1$$, the solution of the non-strict inequality $$\cos x \le a$$ is written as

$\arccos a + 2\pi n \le x \le 2\pi - \arccos a + 2\pi n,\;n \in \mathbb{Z}.$

## Inequalities of the form $$\tan x \gt a,$$ $$\tan x \ge a,$$ $$\tan x \lt a,$$ $$\tan x \le a$$

### Inequality $$\tan x \gt a$$

For any real value of $$a$$, the solution of the strict inequality $$\tan x \gt a$$ has the form

$\arctan a + \pi n \lt x \lt \pi/2 + \pi n,\;n \in \mathbb{Z}.$

### Inequality $$\tan x \ge a$$

For any real value of $$a$$, the solution of the inequality $$\tan x \ge a$$ is expressed in the form

$\arctan a + \pi n \le x \lt \pi/2 + \pi n,\;n \in \mathbb{Z}.$

### Inequality $$\tan x \lt a$$

For any value of $$a$$, the solution of the inequality $$\tan x \lt a$$ is written in the form

$-\pi/2 + \pi n \lt x \lt \arctan a + \pi n,\;n \in \mathbb{Z}.$

### Inequality $$\tan x \le a$$

For any value of $$a$$, the inequality $$\tan x \le a$$ has the following solution:

$-\pi/2 + \pi n \lt x \le \arctan a + \pi n,\;n \in \mathbb{Z}.$

## Inequalities of the form $$\cot x \gt a,$$ $$\cot x \ge a,$$ $$\cot x \lt a,$$ $$\cot x \le a$$

### Inequality $$\cot x \gt a$$

For any value of $$a$$, the solution of the inequality $$\cot x \gt a$$ has the form

$\pi n \lt x \lt \text {arccot } a + \pi n,\;n \in \mathbb{Z}.$

### Inequality $$\cot x \ge a$$

The non-strict inequality $$\cot x \ge a$$ has the similar solution:

$\pi n \lt x \le \text {arccot } a + \pi n,\;n \in \mathbb{Z}.$

### Inequality $$\cot x \lt a$$

For any value of $$a$$, the solution of the inequality $$\cot x \lt a$$ lies on the open interval

$\text {arccot } a + \pi n \lt x \lt \pi + \pi n,\;n \in \mathbb{Z}.$

### Inequality $$\cot x \le a$$

For any value of $$a$$, the solution of the non-strict inequality $$\cot x \le a$$ is in the half-open interval

$\text {arccot } a + \pi n \le x \lt \pi + \pi n,\;n \in \mathbb{Z}.$

## Solved Problems

### Example 1.

Solve the inequality

$\sin x \ge -\frac{1}{2}.$

Solution.

Mark on the $$y-$$axis a point $$y = -\frac{1}{2}.$$ Find the angles corresponding to the given value of the sine:

$\sin x = -\frac{1}{2}, \Rightarrow$
$x_1 = \arcsin{\left({-\frac{1}{2}}\right)} = - \arcsin{\left({\frac{1}{2}}\right)} = -\frac{\pi}{6};$
$x_2 = \pi - {x_1} = \pi - \left({-\frac{\pi}{6}}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6}.$

The solution of inequality $$\sin x \ge -\frac{1}{2}$$ on the unit circle is represented as the sector $$\left[-\frac{\pi}{6}, \frac{7\pi}{6}\right].$$ Do not forget to add periodic terms in the answer:

$-\frac{\pi}{6} + 2\pi n \le x \le \frac{7\pi}{6} + 2\pi n,\;n \in \mathbb{Z}.$

### Example 2.

Solve the inequality

$\cos x \lt -\frac{\sqrt{2}}{2}.$

Solution.

Solving the equation $$\cos x = -\frac{\sqrt{2}}{2}$$ we find:

$x = \pm\arccos \left({-\frac{\sqrt{2}}{2}}\right) + 2\pi n = \pm\left({\pi - \arccos{\frac{\sqrt{2}}{2}}}\right) + 2\pi n = \pm\left({\pi - \frac{\pi}{4}}\right) + 2\pi n = \pm\frac{3\pi}{4} + 2\pi n,\;n \in \mathbb{Z}.$

Notice that the angle $$-\frac{3\pi}{4}$$ can be written as $$\frac{5\pi}{4}.$$ Hence, the solution of the inequality lies within the interval $$\left[{\frac{3\pi}{4}, \frac{5\pi}{4}}\right].$$

$\frac{3\pi}{4} + 2\pi n \lt x \lt \frac{5\pi}{4} + 2\pi n,\;n \in \mathbb{Z}.$

### Example 3.

Solve the inequality

$\tan x \gt -\sqrt{3}.$

Solution.

The solution of the inequality $$\tan x \gt a$$ is written in the form

$\arctan a + \pi n \lt x \lt \pi/2 + \pi n,\;n \in \mathbb{Z}.$

In our case:

$\arctan a = \arctan\left({-\sqrt{3}}\right) = - \arctan\sqrt{3} = -\frac{\pi}{3}.$

Therefore, we have:

$-\frac{\pi}{3} + \pi n \lt x \lt \pi/2 + \pi n,\;n \in \mathbb{Z}.$

### Example 4.

Solve the inequality

$\cot x \ge \frac{1}{\sqrt{3}}.$

Solution.

We know that the solution of the non-strict inequality $$\cot x \ge a$$ is given by

$\pi n \lt x \le \text {arccot } a + \pi n,\;n \in \mathbb{Z}.$

Since

$\text{arccot}\frac{1}{\sqrt{3}} = \frac{\pi}{3},$

we can write

$\pi n \lt x \le \frac{\pi}{3} + \pi n,\;n \in \mathbb{Z}.$