Consider the real-valued function satisfying $2\mathrm{f}(\sin \mathrm{x})+\mathrm{f}(\cos \mathrm{x})=\mathrm{x}$.Then the

Given 2f(sin x) + f (cos x) = x                  (1)
Replacing x by $\frac{\mathrm{\pi }}{2}-\mathrm{x}$, we get
$2\mathrm{f}(\cos \mathrm{x})+\mathrm{f}(\sin \mathrm{x})=\frac{\mathrm{\pi }}{2}-\mathrm{x} \left(2\right)$
Eliminating f(cos x) from (1) and (2), we get
$3\mathrm{f}(\sin \mathrm{x})=3\mathrm{x}-\frac{\mathrm{\pi }}{2}$
or   $\mathrm{f}\left(\mathrm{x}\right)={\sin }^{-1}\mathrm{x}-\frac{\mathrm{\pi }}{6}$
f (x) has the domain [-1, 1].
Also, ${\sin }^{-1} \mathrm{x}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$
or  ${\sin }^{-1} \mathrm{x}-\frac{\mathrm{\pi }}{6}\in \left[-\frac{2\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{3}\right]$