# Systems of Trigonometric Equations

A system of trigonometric equations is a set of equations composed either only from trigonometric equations or from trigonometric and algebraic equations.

In this section, we consider several types of systems of trigonometric equations with two variables $$x$$ and $$y$$ and describe possible ways to solve them.

## Simplest Systems of Equations

It may happen that one of the equations of the system contains trigonometric functions in the unknowns x and y, and the other equation is linear in x and y. In such case, we act in an obvious way: one of the unknowns is expressed from the linear equation and substitute into another equation of the system.

### Example

Solve the system:

$\left\{ \begin{array}{l} x + y = \frac{\pi}{2}\\ \sin x + \sin y = 1 \end{array} \right..$

Solution.

From the first equation, we express $$y$$ in terms of $$x:$$

$y = \frac{\pi}{2} - x,$

and substitute into the second equation:

$\sin x + \sin y = 1, \Rightarrow \sin x + \sin \left({\frac{\pi}{2} - x}\right) = 1, \Rightarrow \sin x + \cos x = 1.$

We got a linear trigonometric equation. Let's solve it using the R method. Multiply both sides of the equation by $$\frac{\sqrt{2}}{2}$$ and convert the expression in the left-hand side into one trig function:

$\frac{\sqrt{2}}{2}\sin x + \frac{\sqrt{2}}{2}\cos x = \frac{\sqrt{2}}{2}, \Rightarrow \cos\frac{\pi}{4}\sin x + \sin\frac{\pi}{4}\cos x = \frac{\sqrt{2}}{2}, \Rightarrow \sin\left({x + \frac{\pi}{4}}\right) = \frac{\sqrt{2}}{2}.$

The solution of this equation is written in the form of two branches:

$x + \frac{\pi}{4} = \frac{\pi}{4} + 2\pi n, \Rightarrow x_1 = 2\pi n,$
$x + \frac{\pi}{4} = \frac{3\pi}{4} + 2\pi n, \Rightarrow x_2 = \frac{\pi}{2} + 2\pi n,$

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

Find now the corresponding values of $$y:$$

$y_1 = \frac{\pi}{2} - x_1 = \frac{\pi}{2} - 2\pi n,$
$y_2 = \frac{\pi}{2} - x_2 = \cancel{\frac{\pi}{2}} - \cancel{\frac{\pi}{2}} - 2\pi n = - 2\pi n, \;n \in \mathbb{Z}.$

The final answer is written as $$\left({x,y}\right)$$ pairs:

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

## Reduction of a Trigonometric System to an Algebraic One

In a number of cases, the trigonometric system can be reduced to a system of algebraic equations by a suitable change of variables.

### Example

Solve the system:

$\left\{ \begin{array}{l} \sin x + \cos y = 1\\ \cos2x - \cos 2y = 1 \end{array} \right..$

Solution.

Remember that

$\cos2x = 1 - 2\sin^2x,\;\cos2y = 2\cos^2y - 1.$

Then we can make the substitution $$u = \sin x,$$ $$v = \cos y$$ which leads to an algebraic system with respect to $$u$$ and $$v:$$

$\left\{ \begin{array}{l} u + v = 1\\ 1 - 2u^2 - \left({2v^2 - 1}\right) = 1 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} u + v = 1\\ 1 - 2u^2 - 2v^2 + 1 = 1 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} u + v = 1\\ u^2 + v^2 = \frac{1}{2} \end{array} \right..$

Express $$v$$ from the first equation: $$v = 1 - u.$$ Substituting $$v$$ into the second equation, we get

$\left\{ \begin{array}{l} v = 1 - u\\ u^2 + \left({1 - u}\right)^2 = \frac{1}{2} \end{array} \right., \Rightarrow \left\{ \begin{array}{l} v = 1 - u\\ u^2 + 1 - 2u + u^2 = \frac{1}{2} \end{array} \right., \Rightarrow \left\{ \begin{array}{l} v = 1 - u\\ 4u^2 - 4u + 1 = 0 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} v = 1 - u\\ \left({2u - 1}\right)^2 = 0 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} v = \frac{1}{2}\\ u = \frac{1}{2} \end{array} \right..$

From here we find $$x$$ and $$y:$$

$\sin x = u = \frac{1}{2}, \Rightarrow x = \left({-1}\right)^n\frac{\pi}{6} + \pi n,\;n\in\mathbb{Z};$
$\cos y = v = \frac{1}{2}, \Rightarrow y = \pm\frac{\pi}{3} + 2\pi k,\;k\in\mathbb{Z}.$

$\left({\left({-1}\right)^n\frac{\pi}{6} + \pi n,\,\pm\frac{\pi}{3} + 2\pi k}\right), \;n,k\in\mathbb{Z}.$

## Addition and Subtraction of Equations

Sometimes a trigonometric system can be simplified by adding or subtracting equations.

### Example

Solve the system:

$\left\{ \begin{array}{l} \sin x \cos y = \frac{1}{4}\\ \cos x \sin y = \frac{3}{4} \end{array} \right..$

Solution.

$\sin x \cos y + \cos x \sin y = \frac{1}{4} + \frac{3}{4},$
$\Rightarrow \sin \left({x + y}\right) = 1.$

Similarly, subtracting the second equation from the first one and using the sine subtraction identity, we have

$\sin x \cos y - \cos x \sin y = \frac{1}{4} - \frac{3}{4},$
$\Rightarrow \sin \left({x - y}\right) = -\frac{1}{2}.$

As a result, we obtain the following equivalent system of equations:

$\left\{ \begin{array}{l} \sin \left({x + y}\right) = 1\\ \sin \left({x - y}\right) = -\frac{1}{2} \end{array} \right..$

The solution of the first equation is written in the form

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

The second equation has two families of solutions:

$x - y = -\frac{\pi}{6} + 2\pi k,$
$x - y = -\frac{5\pi}{6} + 2\pi k,\;k\in\mathbb{Z}.$

Accordingly, we get two solutions:

$\left[ \begin{array}{l} \left\{ \begin{array}{l} x + y = \frac{\pi}{2} + 2\pi n\\ x - y = -\frac{\pi}{6} + 2\pi k \end{array} \right.\\ \left\{ \begin{array}{l} x + y = \frac{\pi}{2} + 2\pi n\\ x - y = -\frac{5\pi}{6} + 2\pi k \end{array} \right. \end{array} \right.,$

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

Solve now each system with respect to $$x$$ and $$y.$$ To do this, add and subtract the equations again:

$\left[ \begin{array}{l} \left\{ \begin{array}{l} 2x = \frac{\pi}{2} -\frac{\pi}{6} + 2\pi\left({n + k}\right)\\ 2y = \frac{\pi}{2} +\frac{\pi}{6} +2\pi\left({n - k}\right) \end{array} \right.\\ \left\{ \begin{array}{l} 2x = \frac{\pi}{2} -\frac{5\pi}{6} + 2\pi\left({n + k}\right)\\ 2y = \frac{\pi}{2} +\frac{5\pi}{6} +2\pi\left({n - k}\right) \end{array} \right. \end{array} \right., \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = \frac{\pi}{6} + \pi\left({n + k}\right)\\ y = \frac{\pi}{3} + \pi\left({n - k}\right) \end{array} \right.\\ \left\{ \begin{array}{l} x = -\frac{\pi}{6} + \pi\left({n + k}\right)\\ y = \frac{2\pi}{3} + \pi\left({n - k}\right) \end{array} \right. \end{array} \right..$

$\left({\frac{\pi}{6} + \pi\left({n + k}\right),\,\frac{\pi}{3} + \pi\left({n - k}\right)}\right),\left({-\frac{\pi}{6} + \pi\left({n + k}\right),\,\frac{2\pi}{3} + \pi\left({n - k}\right)}\right), \;n,k\in\mathbb{Z}.$

## Multiplication and Division of Equations

Sometimes you can come to a solution by multiplying or dividing the equations by each other.

### Example

Solve the system:

$\left\{ \begin{array}{l} \sin x + \sin y = 1\\ \cos x - \cos y = \sqrt{3} \end{array} \right..$

Solution.

Using sum-to-product identities, we write the system as

$\left\{ \begin{array}{l} 2\sin\frac{x+y}{2}\cos\frac{x-y}{2} = 1\\ -2\sin\frac{x+y}{2}\sin\frac{x-y}{2} = \sqrt{3} \end{array} \right..$

Denote $$\alpha = \frac{x+y}{2},$$ $$\beta = \frac{x-y}{2}$$ and rewrite the system in the following form:

$\left\{ \begin{array}{l} 2\sin\alpha\cos\beta = 1\\ -2\sin\alpha\sin\beta = \sqrt{3} \end{array} \right..$

It is clear that $$\sin\alpha \ne 0.$$ Divide the second equation by the first:

$-\tan\beta = \sqrt{3}, \Rightarrow \tan\beta = -\sqrt{3}, \Rightarrow \beta = \arctan\left({-\sqrt{3}}\right) + \pi n = -\frac{\pi}{3} + \pi n, \;n\in\mathbb{Z}.$

The angle $$\beta$$ takes two values on the unit circle. Represent them in separate formulas:

$\beta_1 = -\frac{\pi}{3} + 2\pi n,\; \beta_2 = \frac{2\pi}{3} + 2\pi n,\;n\in\mathbb{Z}.$

Calculate the corresponding values of $$\alpha:$$

$\beta_1 = -\frac{\pi}{3} + 2\pi n, \Rightarrow \cos\beta_1 = \cos\left({-\frac{\pi}{3}}\right) = \frac{1}{2}, \Rightarrow 2\sin\alpha_1\cos\beta_1 = 1, \Rightarrow \sin\alpha_1 = 1, \Rightarrow \alpha_1 = \frac{\pi}{2} + 2\pi k,\;k\in\mathbb{Z}.$
$\beta_2 = \frac{2\pi}{3} + 2\pi n, \Rightarrow \cos\beta_2 = \cos\left({\frac{2\pi}{3}}\right) = -\frac{1}{2}, \Rightarrow 2\sin\alpha_2\cos\beta_2 = 1, \Rightarrow \sin\alpha_2 = -1, \Rightarrow \alpha_2 = -\frac{\pi}{2} + 2\pi k,\;k\in\mathbb{Z}.$

So we get the following solutions for $$\alpha$$ and $$\beta:$$

$\left[ \begin{array}{l} \left\{ \begin{array}{l} \alpha_1 = \frac{\pi}{2} + 2\pi k\\ \beta_1 = -\frac{\pi}{3} + 2\pi n \end{array} \right.\\ \left\{ \begin{array}{l} \alpha_2 = -\frac{\pi}{2} + 2\pi k\\ \beta_2 = \frac{2\pi}{3} + 2\pi n \end{array} \right. \end{array} \right.,\;k,n\in\mathbb{Z}.$

Hence,

$\left[ \begin{array}{l} \left\{ \begin{array}{l} \frac{x+y}{2} = \frac{\pi}{2} + 2\pi k\\ \frac{x-y}{2} = -\frac{\pi}{3} + 2\pi n \end{array} \right.\\ \left\{ \begin{array}{l} \frac{x+y}{2} = -\frac{\pi}{2} + 2\pi k\\ \frac{x-y}{2} = \frac{2\pi}{3} + 2\pi n \end{array} \right. \end{array} \right., \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = \frac{\pi}{2} - \frac{\pi}{3} + 2\pi\left({k + n}\right)\\ y = \frac{\pi}{2} + \frac{\pi}{3} + 2\pi\left({k - n}\right) \end{array} \right.\\ \left\{ \begin{array}{l} x = -\frac{\pi}{2} + \frac{2\pi}{3} + 2\pi\left({k + n}\right)\\ y = -\frac{\pi}{2} - \frac{2\pi}{3} + 2\pi\left({k - n}\right) \end{array} \right. \end{array} \right., \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = \frac{\pi}{6} + 2\pi\left({k + n}\right)\\ y = \frac{5\pi}{6} + 2\pi\left({k - n}\right) \end{array} \right.\\ \left\{ \begin{array}{l} x = \frac{\pi}{6} + 2\pi\left({k + n}\right)\\ y = -\frac{7\pi}{6} + 2\pi\left({k - n}\right) \end{array} \right. \end{array} \right.,$

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

$\left({\frac{\pi}{6} + 2\pi\left({k + n}\right),\,\frac{5\pi}{6} + 2\pi\left({k - n}\right)}\right),\left({\frac{\pi}{6} + 2\pi\left({k + n}\right),\,-\frac{7\pi}{6} + 2\pi\left({k - n}\right)}\right), \;k,n\in\mathbb{Z}.$