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 \(y:\)
\[\sin x = u = \frac{1}{2}, \Rightarrow x = \left({-1}\right)^n\frac{\pi}{6} + \pi n,\;n\in\mathbb{Z};\]
\[\cos y = v = \frac{1}{2}, \Rightarrow y = \pm\frac{\pi}{3} + 2\pi k,\;k\in\mathbb{Z}.\]
The answer is given by
\[\left({\left({-1}\right)^n\frac{\pi}{6} + \pi n,\,\pm\frac{\pi}{3} + 2\pi k}\right), \;n,k\in\mathbb{Z}.\]
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.
Adding two equations and using the sine addition identity, we get
\[\sin x \cos y + \cos x \sin y = \frac{1}{4} + \frac{3}{4},\]
\[\Rightarrow \sin \left({x + y}\right) = 1.\]
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:
\[\left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x = \frac{\pi}{2} -\frac{\pi}{6} + 2\pi\left({n + k}\right)\\
2y = \frac{\pi}{2} +\frac{\pi}{6} +2\pi\left({n - k}\right)
\end{array} \right.\\
\left\{ \begin{array}{l}
2x = \frac{\pi}{2} -\frac{5\pi}{6} + 2\pi\left({n + k}\right)\\
2y = \frac{\pi}{2} +\frac{5\pi}{6} +2\pi\left({n - k}\right)
\end{array} \right.
\end{array} \right., \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = \frac{\pi}{6} + \pi\left({n + k}\right)\\
y = \frac{\pi}{3} + \pi\left({n - k}\right)
\end{array} \right.\\
\left\{ \begin{array}{l}
x = -\frac{\pi}{6} + \pi\left({n + k}\right)\\
y = \frac{2\pi}{3} + \pi\left({n - k}\right)
\end{array} \right.
\end{array} \right..\]
The answer is written as
\[\left({\frac{\pi}{6} + \pi\left({n + k}\right),\,\frac{\pi}{3} + \pi\left({n - k}\right)}\right),\left({-\frac{\pi}{6} + \pi\left({n + k}\right),\,\frac{2\pi}{3} + \pi\left({n - k}\right)}\right), \;n,k\in\mathbb{Z}.\]
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 \(k,n \in \mathbb{Z}.\)
The answer is given by
\[\left({\frac{\pi}{6} + 2\pi\left({k + n}\right),\,\frac{5\pi}{6} + 2\pi\left({k - n}\right)}\right),\left({\frac{\pi}{6} + 2\pi\left({k + n}\right),\,-\frac{7\pi}{6} + 2\pi\left({k - n}\right)}\right), \;k,n\in\mathbb{Z}.\]
