The number of solutions of the trigonometric equations 1+cosx.cos5x=sin2x in [0,2π] is

cosx.cos5x+cos2x=0 ⇒cosx=0 (or) cos5x+cosx=0⇒cosx=0, 2cos3x.cos2x=0⇒x=(2n+1)π2, 3x=(2n+1)π2, 2x=(2n+1)π2x=(2n+1)π2, x=(2n+1)π6, x=(2n+1)π4∴Set of solutions is {π2,3π2,π6,5π6,7π6,11π6,π4,3π4,5π4,7π4} ∵x∈0,2π