The problem says:
What is $f'(0)$, given that
$f\left(\sin x −\frac{\sqrt 3}{2}\right) = f(3x − \pi) + 3x − \pi$, $x \in [−\pi/2, \pi/2]$.
So I called $g(x) = \sin x −\dfrac{\sqrt 3}{2}$ and $h(x)=3x − \pi$.
Since $f(g(x)-h(x))=3x − \pi$, I called $g(x)-h(x) = j(x) = \sin x −\frac{\sqrt 3}{2} - 3x +\pi$
I imagined that I should find the function $f$, by finding the inverse of $j(x)$ and doing $(f \circ j \circ j^{-1})(x)$ but I discovered that it is too much difficult.
I guess that's not the best way.
What should I do?
Given: $$f(\sin x - \frac{\sqrt 3}{2}) = f(3x-\pi) + 3x - \pi$$ Taking derivatives on both sides w.r.t. $x$, we get, $$f'(\sin x - \frac{\sqrt 3}{2}).\cos x = 3.f'(3x-\pi) + 3$$ Put $x=\frac{\pi}{3}$, we get, $$f'(\sin \frac{\pi}{3} - \frac{\sqrt 3}{2}).\cos \frac{\pi}{3} = 3.f'(3.\frac{\pi}{3}-\pi) + 3$$ $$f'(0).\frac{1}{2} = 3.f'(0) + 3$$ Solving for $f'(0)$, we get, $$f'(0)=-\frac{6}{5}$$