Solving trigonometric equations

In order to solve a trigonometric equation we will follow these steps:

1) We develop the expressions until we obtain only one trigonometric expression equaling to a number.

2) We will obtain one of the following equalities: $\begin{array}{rcl} \sin u & = & a \\ \cos u & = & b \\ \tan u & = & c \end{array}$ where $u$ is a function of $x$ .

3) We solve each of them by taking the arc of the corresponding functions in the two sides of the equations:

$\sin u = a \Rightarrow \arcsin (\sin u) = \arcsin a \Rightarrow$

$u = {\begin{cases} \arcsin a + 2 k \cdot π \\ (π - \arcsin a) + 2 k \cdot π \end{cases}, k \in Z$

$\cos u = b \Rightarrow \arccos (\cos u) = \arccos b \Rightarrow$

$u = {\begin{cases} \arccos b + 2 k \cdot π \\ (2 π - \arccos b) + 2 k \cdot π \end{cases}, k \in Z$

$\tan u = c \Rightarrow \arctan (\tan u) = \arctan c \Rightarrow u = \arctan c + π \cdot k$

4) Once we have $u$ , we isolate $x$ .

Example

Let's solve the following trigonometric equation: $\sin^{2} x - \cos^{2} x = \frac{1}{2}$

First, we isolate $\sin^{2} x$ : $\sin^{2} x = \frac{1}{2} + \cos^{2} x$

From the relation: $\sin^{2} x + \cos^{2} x = 1 \Rightarrow \cos^{2} x = 1 - \sin^{2} x$ whereby we substitute in our equation: $s i n^{2} x = \frac{1}{2} + \cos^{2} x = \frac{1}{2} + 1 - \sin^{2} x = \frac{3}{2} - s i n^{2} x \Rightarrow 2 \sin^{2} x = \frac{3}{2} \Rightarrow$ $\Rightarrow \sin^{2} x = \frac{3}{4} \Rightarrow \sin x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$ Now we have already managed to obtain a trigonometric ratio which equals to a number.

We apply now the relation 3.i in the two possible situations:

Case (a): $\sin x = \frac{\sqrt{3}}{2} \Rightarrow x = {\begin{cases} \frac{π}{3} + 2 π \cdot k \\ π - \frac{π}{3} + 2 π \cdot k = \frac{2 π}{3} + 2 π \cdot k \end{cases}, k \in Z$

Case (b): $\sin x = - \frac{\sqrt{3}}{2} \Rightarrow x = {\begin{cases} - \frac{π}{3} + 2 π \cdot k \\ π + \frac{π}{3} + 2 π \cdot k = \frac{4 π}{3} + 2 π \cdot k \end{cases}, k \in Z$ So we obtain the following solution: $x = {\begin{cases} \frac{π}{3} + 2 π \cdot k \\ \frac{2 π}{3} + 2 π \cdot k \\ - \frac{π}{3} + 2 π \cdot k \\ \frac{4 π}{3} + 2 π \cdot k \end{cases}, k \in Z$