The number of solutions of the equation 2(sin−1x)2−5sin−1x+2=0 is
0
1
2
3
We have 2sin-1x2-5sin-1x+2=0 ⇒2sin-1x-1sin-1x-2=0 ⇒2sin-1x-1=0 ∵-π2≤sin-1x≤π2⇒sin-1x-2≠0 ⇒x=sin12 is the only solution of the equation