The number of real roots of the equation cosec⁡θ+sec⁡θ−15=0 lying in [0,π] is

The given equation becomes sin⁡θ+cos⁡θ=15sin⁡θcos⁡θ Squaring, 1+sin⁡2θ=15sin2⁡θcos2⁡θ1+sin⁡2θ=154sin2⁡2θ⇒15sin2⁡2θ−4sin⁡2θ−4=0⇒ sin⁡2θ=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.