TITLE: What are the ways to solve trig equations of the form $\sin(f(x)) = \cos(g(x))$?
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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 $$f(x)-g(x)=n\pi-\frac\pi2$$