help!!

for, x = pi/2, 2f(1) + f(0) = pi/2

for, x = 0 , 2f(0) + f(1) = 0

for, x = -pi/2 , 2f(-1) + f(0) = -pi/2

we get, f(1) = pi/3

f(0) = -pi/6

f(-1) = -5pi/12

which gives f(x) = sin^-1 x - x/6

so, domain -1 <= x =< 1

range pi/3 to -5pi/12

ishan u seem to hav made a very small error it seems

f(x)=sin^{-1}x - \frac{\pi }{6}, seems correct

\frac{-\pi }{2} \leq sin^{-1}x\leq \frac{\pi }{2}

\frac{-\pi }{2}-\frac{\pi }{6} \leq (sin^{-1}x-\frac{\pi }{6})\leq \frac{\pi }{2}-\frac{\pi }{6}

\frac{-2\pi }{3}\leq (sin^{-1}x-\frac{\pi }{6})\leq \frac{\pi }{3}

wich is the ans !!!! thanx ishan !!

well akari gav a good method to find f(x)

write first functional eqn as it is ,

then put x = pie/2 - x

get the second functional equation ,

from these 2 functional eqns ,
get f(sinx ) , and put x = sin^-1x

then u get f(x)