The correct option is C 2
√2√(sin2x)=√2sin−1sinx,−π<x≤π
(i)0≤x≤π2, then sin x=sin−1sinx=x.
X = 0 is obviously a solution and in 0<x≤π2,sinx<x.
(ii)π2<x≤π, we get
sinx=sin−1sin(π−x)=π−x
x=π is only solution in this case.
(iii)−π2≤x<0,−sinx=x which has no solution in this interval.
(iv)−π<x<−π2⇒−sinx=−π−x⇒sinx=π+x
Which has no solution in the given interval.
Hence, there are only two solutions x=0,π