Is there a general approach to solving trigonometric equations? We've been solving equations in school that mainly include forms of sin, cos and tan but usually I don't know how to approach them. It's not like solving a quadratic where there's just a solution formula.
 A: As user85667 remarks, if the equation only involves trig functions of some variable $x$ (or possibly its integer multiples), then the equation is in principle equivalent to a polynomial equation. However writing this equation down and solving it may be a very non-trivial task.
If the equation only involves $\sin$, $\cos$, and $\tan$, all of them of a single argument $x$, you should look for a way to express everything in terms of just one of these. For example, we have
$$
\sin x = \pm\frac{\tan x}{\sqrt{1+ \tan^2 x}}, ~~ \cos x = \pm \frac{1}{\sqrt{1+ \tan^2 x}},
$$
so this would enable you to express everything in terms of $\tan x$, and then solve for the new variable $u = \tan x$.
However this means you may end up with a lot of square roots. Another way (taking a cue from the book Integration by R.P. Gillespie, p. 23 in the sixth edition) is to take $u=\tan \frac{x}{2}$ instead, because we then have
$$
\sin x = \frac{2u}{1+u^2}, ~~ \cos x = \frac{1-u^2}{1+u^2}, ~~\tan x = \frac{2u}{1-u^2}.
$$
This will get you a polynomial equation in $u=\tan\frac{x}{2}$, which may or may not be easily solvable. Depending on cases, there may be a more clever substitution available.
A: It's very useful to know Euler's Formula
$$e^{i\theta} = \cos \theta + i \sin\theta$$
From this (and its conjugate $e^{-i\theta} = \cos \theta - i \sin\theta$), you can derive the identities:
$$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$$
$$\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$
And from these, you can derive pretty much any trig identity that you need.  For example, have you forgotten what $\sin\alpha\sin\beta$ is equivalent to?  Then just do
$$\sin\alpha\sin\beta = (\frac{e^{i\alpha} - e^{-i\alpha}}{2i})(\frac{e^{i\beta} - e^{-i\beta}}{2i})$$
$$= \frac{(e^{i\alpha} - e^{-i\alpha})(e^{i\beta} - e^{-i\beta})}{-4}$$
$$= \frac{e^{i\alpha}e^{i\beta} - e^{i\alpha}e^{-i\beta} - e^{-i\alpha}e^{i\beta} + e^{-i\alpha}e^{-i\beta}}{-4}$$
$$= \frac{e^{i(\alpha + \beta)} - e^{i(\alpha - \beta)} - e^{i(\beta-\alpha)} + e^{-i(\alpha + \beta)}}{-4}$$
$$= \frac{e^{i(\alpha + \beta)}+ e^{-i(\alpha + \beta)} - e^{i(\alpha - \beta)} - e^{-i(\alpha-\beta)} }{-4}$$
$$= \frac{2\cos(\alpha+\beta) - 2\cos(\alpha-\beta)}{-4}$$
$$= \frac{\cos(\alpha-\beta) - \cos(\alpha+\beta)}{2}$$
