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?

Otherwise  as you ask could be as follows (although it is almost the same).
You have  $$sin (f(x))-sin (\frac\pi2-g(x))=0$$ so from the identity
$$sin(a)-sin(b)=2sin\frac {a-b}{2}cos\frac{a+b}{2}$$ 
it follows $$2sin\frac{(f(x)+g(x)- \frac{\pi}{2})}{2}cos\frac{(f(x)-g(x)+\frac{\pi}{2})}{2}=0$$ hence $$f(x)+g(x)=2n\pi+\frac\pi2$$ and $$f(x)-g(x)=n\pi-\frac\pi2$$
