Solve equations in a given interval and write general solutions using $\sin\theta = \sin\alpha \Rightarrow \theta = n\pi + (-1)^n\alpha$, $\cos\theta = \cos\alpha \Rightarrow \theta = 2n\pi \pm \alpha$ and $\tan\theta = \tan\alpha \Rightarrow \theta = n\pi + \alpha$.
$\sin x = \dfrac12$ is true for $x = \dfrac{\pi}{6}$, but also for $\dfrac{5\pi}{6}$, $\dfrac{13\pi}{6}$, $-\dfrac{7\pi}{6}$, ... because the functions repeat. A question either restricts $x$ to an interval, such as $0 \le x \le 2\pi$ (then list all solutions in it), or asks for the general solution, a formula with an integer $n$ that gives every solution.
Sine. $\sin\theta = \sin\alpha$ means the points $\theta$ and $\alpha$ have the same height on the unit circle, so $\theta = \alpha$ or $\theta = \pi - \alpha$, each plus any multiple of $2\pi$. Both families are written as one formula: $$\sin\theta = \sin\alpha \iff \theta = n\pi + (-1)^n\alpha.$$ (For even $n$ this is $\alpha$ plus a multiple of $2\pi$; for odd $n$ it is $\pi - \alpha$ plus a multiple of $2\pi$.)
Cosine. Equal cosines mean equal horizontal positions, so $\theta = \pm\alpha$ plus a multiple of $2\pi$: $$\cos\theta = \cos\alpha \iff \theta = 2n\pi \pm \alpha.$$
Tangent. The tangent repeats every $\pi$: $$\tan\theta = \tan\alpha \iff \theta = n\pi + \alpha.$$
In each case $n$ is any integer, and $\alpha$ is any one convenient solution, usually the principal value (the one in the middle of the standard range).
Example: Find the general solution of $\sin x = \dfrac{\sqrt3}{2}$. Since $\dfrac{\sqrt3}{2} = \sin\dfrac{\pi}{3}$, $x = n\pi + (-1)^n\dfrac{\pi}{3}$, $n \in \mathbb{Z}$. For $n = 0, 1, 2$ this gives $\dfrac{\pi}{3}$, $\dfrac{2\pi}{3}$, $\dfrac{7\pi}{3}$.
Example: Find the general solution of $\cos 3x = \dfrac12$. $\cos 3x = \cos\dfrac{\pi}{3}$, so $3x = 2n\pi \pm \dfrac{\pi}{3}$, hence $x = \dfrac{2n\pi}{3} \pm \dfrac{\pi}{9}$, $n \in \mathbb{Z}$. Notice that the whole $3x$ is the angle, and we divide everything, including the $2n\pi$, by $3$.
Example: Solve $\tan x = -1$ for $0^\circ \le x \le 360^\circ$, and give the general solution in radians. The reference angle is $45^\circ$ and tangent is negative in the second and fourth quadrants: $x = 135^\circ$ or $315^\circ$. In radians, $\tan x = \tan\left(-\dfrac{\pi}{4}\right)$, so $x = n\pi - \dfrac{\pi}{4}$.
Example: Solve $2\sin^2 x + \sin x - 1 = 0$ for $0 \le x \le 2\pi$. Factorise: $(2\sin x - 1)(\sin x + 1) = 0$. $\sin x = \dfrac12$ gives $x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$; $\sin x = -1$ gives $x = \dfrac{3\pi}{2}$.
Example: Solve $\sin 2x = \cos x$ for $0 \le x \le 2\pi$. $2\sin x\cos x - \cos x = 0$, so $\cos x(2\sin x - 1) = 0$. $\cos x = 0$ gives $x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$; $\sin x = \dfrac12$ gives $x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$. Four solutions. (Dividing by $\cos x$ at the start would have lost the first two.)
Exam tip: In "find the general solution" questions, examiners mark the formula and its form: write $n \in \mathbb{Z}$, and put the $2n\pi$ or $n\pi$ in the right place. In interval questions, count the solutions you expect (for example a quadratic in $\sin x$ over $[0, 2\pi]$ usually gives up to four). When the interval is in degrees, give answers in degrees; when it has $\pi$, give exact multiples of $\pi$ if possible.