We have, 2sin2θ−3cos2θ=sinθcosθ Dividing both sides by cos2θ, we get cosθ≠0 ⇒2tan2θ−3=tanθ ⇒2tan2θ−tanθ−3=0 ⇒(2tanθ−3)(tanθ+1)=0 ∵tanθ=
3
2
and tanθ=−1 Clearly, in the interval (−π,π)tanθ=
3
2
gives two solution and tanθ=−1 gives two solution ∴ Total number of solution =4
