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

Solution:

The given equation becomes $\sin θ + \cos θ = \sqrt{15} \sin θcos θ$
$Squaring, 1 + \sin 2 θ = 15 \sin^{2} {θcos}^{2} θ$
$\begin{array}{l} 1 + \sin 2 θ = \frac{15}{4} \sin^{2} 2 θ \\ \Rightarrow 15 \sin^{2} 2 θ - 4 \sin 2 θ - 4 = 0 \\ \Rightarrow \sin 2 θ = \frac{2}{3}, - \frac{2}{5} \end{array}$
As $θ$ varies from 0 to $π$ ,2 $θ$ varies from $θ$ to 2 $π$ .
Corresponding to $\sin 2 θ = \frac{2}{3}$ ,there are two solutions.
Corresponding to $\sin 2 θ = - \frac{2}{5}$ ,there are two solutions.

The number of real roots of the equation $cosec θ + \sec θ - \sqrt{15} = 0 lying in [0, π]$ is

Solution:

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