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:
$$\sin u=a \Rightarrow \arcsin (\sin u)=\arcsin a \Rightarrow $$
$$$u=\left\{\begin{array}{l} \arcsin a+ 2k\cdot \pi \\ (\pi-\arcsin a)+2k\cdot \pi \end{array}\right., k \in \mathbb{Z} $$$
$$\cos u=b \Rightarrow \arccos (\cos u)=\arccos b \Rightarrow$$
$$$u=\left\{\begin{array}{l} \arccos b+ 2k\cdot \pi \\ (2\pi-\arccos b)+2k\cdot \pi \end{array}\right., k \in \mathbb{Z}$$$
$$\tan u = c \Rightarrow \arctan (\tan u)=\arctan c \Rightarrow u=\arctan c+ \pi \cdot k$$
4) Once we have $$u$$, we isolate $$x$$.
Let's solve the following trigonometric equation: $$$\sin ^2x- \cos ^2x=\displaystyle \frac{1}{2}$$$
First, we isolate $$\sin^2x$$: $$$\sin ^2x=\displaystyle \frac{1}{2}+\cos ^2 x$$$
From the relation: $$$\sin^2x+\cos^2x=1 \Rightarrow \cos^2x=1-\sin ^2x$$$ whereby we substitute in our equation: $$$\displaystyle sin^2x=\frac{1}{2}+\cos^2x=\frac{1}{2}+1-\sin^2x=\frac{3}{2}-sin^2x \Rightarrow 2\sin^2x=\frac{3}{2} \Rightarrow$$$ $$$\Rightarrow \sin^2x=\frac{3}{4} \Rightarrow \sin x=\pm \sqrt {\frac{3}{4}}=\pm \frac{\sqrt{3}}{2}$$$ 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): $$$\sin x=\displaystyle \frac{\sqrt{3}}{2} \Rightarrow x= \left\{\begin{array} {l} \frac{\pi}{3}+2\pi \cdot k \\ \pi-\frac{\pi}{3}+2\pi \cdot k=\frac{2\pi}{3}+2\pi\cdot k\end{array}\right. , k \in \mathbb{Z}$$$
Case (b): $$$\sin x=\displaystyle -\frac{\sqrt{3}}{2} \Rightarrow x= \left\{\begin{array} {l} -\frac{\pi}{3}+2\pi \cdot k \\ \pi+\frac{\pi}{3}+2\pi \cdot k=\frac{4\pi}{3}+2\pi\cdot k\end{array}\right. , k \in \mathbb{Z}$$$ So we obtain the following solution: $$$x=\left\{\begin{array}{l} \frac{\pi}{3}+2\pi \cdot k \\ \frac{2\pi}{3}+2\pi\cdot k \\ -\frac{\pi}{3}+2\pi \cdot k \\ \frac{4\pi}{3}+2\pi\cdot k\end{array}\right., k \in \mathbb{Z}$$$