The number of real roots of the equation cosecθ+secθ−15=0 lying in [0,π] is
6
8
4
0
The given equation becomes sinθ+cosθ=15sinθcosθ Squaring, 1+sin2θ=15sin2θcos2θ1+sin2θ=154sin22θ⇒15sin22θ−4sin2θ−4=0⇒ sin2θ=23,−25As θ varies from 0 to π,2θ varies from θ to 2π.Corresponding to sin 2θ=23,there are two solutions.Corresponding to sin 2θ=−25,there are two solutions.