It is given that, 2f(sinx)+f(cosx)=x ....(i) by replacing x by 2π−x, we get 2f(cosx)+f(sinx)=2π−x...(ii) from Eqs. (i) and (ii), we get 3f(sinx)=3x−2π⇒3f(x)=3sin−1x−2π( on replacing x by sin−1x) Now, on differentiating both sides w.r.t. ′X′, we get