Proving the solutions of trigonometric functions

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If $f(x) =a^{\cos x}$ and $g(x)=(\sin x)^a$, $a \in N$, then prove that $f(x) = g(x)$ for infinitely many values of x.

My Attempt to prove this: If $f(x) = g(x)$, $$a^{\cos x} = (\sin x)^a$$ Taking Logs on both sides.. $$\log_a a^{\cos x} = \log_a (\sin x)^a$$ $$\cos x = a\log_a \sin x$$

This is just my thought process and I am pretty sure that it is wrong. But it may be something along these lines. I do not know how to start thinking about this problem. Please Help!

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For $x = \frac{\pi}{2} + 2k\pi$, with $k \in \mathbb{Z}$, you have $$f(x)=a^{\cos(x)}=a^0=1$$

and

$$g(x)=(\sin(x))^a=1^a=1$$

So here are your infinite values of $x$ where $f(x)=g(x)$.