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].