# Basic Trigonometric Equations

Angles (arguments of functions): x, x1, x2
Set of integers: Ζ
Integer: n
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 equation involving trigonometric functions of an unknown angle is called a trigonometric equation.

Basic trigonometric equations have the form

$\sin x = a,\;\cos x = a,\;\tan x = a,\;\cot x = a,$

here $$x$$ is an unknown, $$a$$ is any real number.

## Equation $$\sin x = a$$

If $$\left| a \right| \gt 1$$, the equation $$\sin x = a$$ has no solutions.

If $$\left| a \right| \le 1,$$ the general solution of the equation $$\sin x = a$$ is written as

This formula contains two branches of solutions:

${x_1} = \arcsin a + 2\pi n,\;{x_2} = \pi - \arcsin a + 2\pi n,\;n \in \mathbb{Z}.$

The solutions of a trigonometric equation that lie in the interval $$\left[ {0,2\pi } \right)$$ are called principal solutions. The principal solutions for the equation $$\sin x = a$$ are

$\arcsin a,\;\pi - \arcsin a.$

In the simple case $$\sin x = 1$$ the general solution has the form

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

Similarly, the solution of the equation $$\sin x = -1$$ is given by

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

Case $$\sin x = 0$$ (zeroes of the sine):

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

## Equation $$\cos x = a$$

If $$\left| a \right| \gt 1,$$ the equation $$\cos x = a$$ has no solutions.

If $$\left| a \right| \le 1,$$ the general solution of the equation $$\cos x = a$$ has the form

This formula includes two sets of solutions:

${x_1} = \arccos a + 2\pi n,\; {x_2} = -\arccos a + 2\pi n,\; n \in \mathbb{Z}.$

In the case $$\cos x = 1$$, the solution is written as

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

Case $$\cos x = -1:$$

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

Case $$\cos x = 0$$ (zeroes of the cosine):

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

## Equation $$\tan x = a$$

For any value of $$a$$, the general solution of the equation $$\tan x = a$$ has the form

Case $$\tan x = 0$$ (zeroes of the tangent):

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

## Equation $$\cot x = a$$

For any value of $$a$$, the general solution of the trigonometric equation $$\cot x = a$$ is written as

Case $$\cot x = 0$$ (zeroes of the cotangent):

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

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Solve the equation

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

### Example 2

Solve the equation

$\cos \left( {x + \frac{\pi }{3}} \right) = - 1.$

### Example 3

Find the general solution of the equation

$\sqrt 3 \sin x = \cos x.$

### Example 4

Find the principal solutions of the equation

$\cot \left( {2x + \frac{\pi }{4}} \right) = - 1.$

### Example 5

Solve the equation

${\cos ^2}x = \frac{1}{2}.$

### Example 6

Solve the equation

$\tan x = \cot x.$

### Example 1.

Solve the equation

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

Solution.

The general solution of the equation has the form

$x = {\left( { - 1} \right)^n}\arcsin \left( { - \frac{1}{2}} \right) + \pi n,\;n \in \mathbb{Z}.$

The inverse sine function is odd, so that

$\arcsin \left( { - \frac{1}{2}} \right) = - \arcsin \frac{1}{2} = - \frac{\pi }{6}.$

Hence, the general solution is written as

$x = {\left( { - 1} \right)^n}\left( { - \frac{\pi }{6}} \right) + \pi n = {\left( { - 1} \right)^{n + 1}}\frac{\pi }{6} + \pi n,\;n \in \mathbb{Z}.$

The principal solutions on the interval $$\left[ {0,2\pi } \right)$$ are given by

$\begin{array}{*{20}{l}} {n = 1:}&{x_1} = \frac{\pi }{6} + \pi = \frac{{7\pi }}{6}\\ {n = 2:}&{x_2} = - \frac{\pi }{6} + 2\pi = \frac{{11\pi }}{6} \end{array}$

### Example 2.

Solve the equation

$\cos \left( {x + \frac{\pi }{3}} \right) = - 1.$

Solution.

In this special case, the general solution is given by

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

Solve it for $$x:$$

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

The principal solution contains one value:

$n = 0:\;{x_0} = \frac{{2\pi }}{3}.$

### Example 3.

Find the general solution of the equation

$\sqrt 3 \sin x = \cos x.$

Solution.

We rewrite the equation in the form

$\sqrt 3 \sin x - \cos x = 0$

and divide both sides by $$\cos x.$$ Note that $$\cos x \ne 0.$$ Indeed, if $$\cos x = 0,$$ then

$\sin x = \pm \sqrt {1 - {{\cos }^2}x} = \pm 1 \ne 0,$

that is, $$\cos x = 0$$ cannot be a solution of the equation. So we have

$\frac{{\sqrt 3 \sin x}}{{\cos x}} - \frac{{\cancel{{\cos x}}}}{{\cancel{{\cos x}}}} = 0, \Rightarrow \sqrt 3 \tan x - 1 = 0, \Rightarrow \tan x = \frac{1}{{\sqrt 3 }}.$

The general solution is given by

$x = \arctan \frac{1}{{\sqrt 3 }} + \pi n = \frac{\pi }{6} + \pi n,\;n \in \mathbb{Z}.$

### Example 4.

Find the principal solutions of the equation

$\cot \left( {2x + \frac{\pi }{4}} \right) = - 1.$

Solution.

First we find the general solution. Given that $$\text{arccot}\left( { - a} \right) = \pi - \text{arccot } a,$$ we have

$2x + \frac{\pi }{4} = \text{arccot} \left( { - 1} \right) + \pi n, \Rightarrow 2x + \frac{\pi }{4} = \pi - \text{arccot } 1 + \pi n, \Rightarrow 2x + \frac{\pi }{4} = \pi - \frac{\pi }{4} + \pi n, \Rightarrow 2x = \frac{\pi }{2} + \pi n, \Rightarrow x = \frac{\pi }{4} + \frac{{\pi n}}{2},$

where $$n \in \mathbb{Z}.$$

The principal values lie in the interval $$\left[ {0,2\pi } \right).$$ Hence, our principal solutions will be

$\begin{array}{*{20}{l}} {n = 0:}&{{x_0} = \frac{\pi }{4}}\\ {n = 1:}&{{x_1} = \frac{\pi }{4} + \frac{\pi }{2} = \frac{{3\pi }}{4}}\\ {n = 2:}&{{x_2} = \frac{\pi }{4} + \pi = \frac{{5\pi }}{4}}\\ {n = 3:}&{{x_3} = \frac{\pi }{4} + \frac{3\pi }{2} = \frac{{7\pi }}{4}} \end{array}$

### Example 5.

Solve the equation

${\cos ^2}x = \frac{1}{2}.$

Solution.

This equation has two solutions:

$\cos x = \frac{1}{2}, \Rightarrow {x_1} = \pm \arccos \frac{1}{2} + 2\pi n,\;n \in \mathbb{Z}.$
$\cos x = - \frac{1}{2}, \Rightarrow {x_2} = \pm \arccos \left( { - \frac{1}{2}} \right) + 2\pi k,\;k \in \mathbb{Z}.$

Substitute the values of inverse cosine:

$\arccos \frac{1}{2} = \frac{\pi }{3},\;\;\arccos \left( { - \frac{1}{2}} \right) = \pi - \arccos \frac{1}{2} = \pi - \frac{\pi }{3} = \frac{{2\pi }}{3}.$

So the general solution is given by

${x_1} = \pm \frac{\pi }{3} + 2\pi n,\;\;{x_2} = \pm \frac{{2\pi }}{3} + 2\pi k,$

where $$n, k \in \mathbb{Z}.$$

Respectively, the principal solutions of the equation are

$x = \frac{\pi }{3},\frac{{2\pi }}{3},\frac{{4\pi }}{3},\frac{{5\pi }}{3}.$

### Example 6.

Solve the equation

$\tan x = \cot x.$

Solution.

We rewrite the equation as follows:

$\tan x = \cot x, \Rightarrow \tan - \frac{1}{{\tan x}} = 0, \Rightarrow \frac{{{{\tan }^2}x - 1}}{{\tan x}} = 0, \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{{\tan }^2}x = 1}\\ {\tan x \ne 0} \end{array}} \right., \Rightarrow \tan x = \pm 1.$

We have obtained two equations. The first equation $$\tan x = 1$$ has the following solution:

$\tan x = 1, \Rightarrow {x_1} = \arctan 1 + \pi n = \frac{\pi }{4} + \pi n,\;n \in \mathbb{Z}.$

The second equation has a solution in the form

$\tan x = - 1, \Rightarrow {x_2} = \arctan \left( { - 1} \right) + \pi k = - \arctan 1 + \pi k = - \frac{\pi }{4} + \pi k,\;k \in \mathbb{Z}.$

We can merge both solutions and express them with one formula:

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

The principal values are given by

$x = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4}.$