Given that:

2*f(sin x) + f(cos x) = x

Find f'(x)

Lets start with a trick: substitute x -> pi/2-x.

This simplifies to 2*f(cos x) + f(sin x) = pi/2 - x.

Next up take a linear combination of the two equations 2*[2*f(sin x) + f(cos x) = x] - [2*f(cos x) + f(sin x) = pi/2 - x].

This yields f(sin x) = x - pi/6.

Then a simple substitution of x -> arcsin(x) gives us __f(x) = arcsin(x) - pi/6__. Then take the derivative to get **f'(x) = 1/sqrt[1-x^2]**.