<s>[INST] What are the ways to solve trig equations of the form $\sin(f(x)) = \cos(g(x))$? if I have the following trig equation:
$$\sin(10x) = \cos(2x)$$
I take the following steps to solve it:


*

*I rewrite $\cos(2x)$ as $\sin\left(\frac{\pi}{2} + 2x\right)$ or as $\sin\left(\frac{\pi}{2} - 2x\right)$ cause $\sin\left(\frac{\pi}{2} - a\right) = \sin\left(\frac{\pi}{2} + a\right) = \cos(a)$;

*Let's say that I have chosen $\sin\left(\frac{\pi}{2} + 2x\right)$, the equation becomes:


$$\sin(10x) = \sin\left(\frac{\pi}{2} + 2x\right).$$


*

*Then, I know that:


$$\sin(f(x)) = \sin(g(x)) \Leftrightarrow f(x) = g(x) + 2\pi n, n \in \mathbb{Z} \lor f(x) = (\pi - g(x)) + 2\pi k, k \in \mathbb{Z}$$
This means that (for $f(x) = 10x$ and $g(x) = \frac{\pi}{2} + 2x$):
$$\sin(10x) = \cos(2x) \Leftrightarrow 10x = \frac{\pi}{2} + 2x + 2\pi n, n \in \mathbb{Z} \lor 10x = (\pi - (\frac{\pi}{2} + 2x)) + 2\pi k, k \in \mathbb{Z}$$
Solving, I get the following results:
$$x_{1} = \frac{\pi}{16} + \frac{\pi}{4}n,\,\,x_{2} = \frac{\pi}{24} + \frac{\pi}{6}k,\,\,\,\,n,k \in \mathbb{Z}$$
Now, are there any other methods for solving such equations or could this one be just fine?
 [/INST] You may prefer to transform the sine into cosine:
$$
\cos\left(\frac{\pi}{2}-10x\right)=\cos(2x)
$$
This splits into two:
$$
\frac{\pi}{2}-10x=2x+2k\pi
$$
or
$$
\frac{\pi}{2}-10x=-2x+2k\pi
$$
The trick is that $\cos\alpha=\cos\beta$ if and only if $\alpha=\beta+2k\pi$ or $\alpha=-\beta+2k\pi$ (with integer $k$).
</s>