Question

Find number of real roots of the equation $\sec \theta +\text{cosec}\theta =\sqrt{15}$ lying between $0$ and $\pi$.

Answer 1

$\sec \theta +\text{cosec}\theta =\sqrt{15}$

$\Rightarrow \sin \theta +\cos \theta =\sqrt{15}.\sin \theta .\cos \theta$

Squaring both sides we get,

$\Rightarrow 1+\sin 2\theta =15(\sin \theta .\cos \theta {)}^2$

$\Rightarrow 4+4\sin 2\theta =15(\sin 2\theta {)}^2$

Solving above quadratic equation we get

$\sin 2\theta =\frac{4+\sqrt{16\times 15+16}}{2\times 15}$ or

$\sin 2\theta =\frac{4-\sqrt{16\times 15+16}}{2\times 15}$

$\Rightarrow \sin 2\theta =\frac{2}{3}$ or $\frac{-2}{5}$

$\theta ={20.90}^{o}or-{11.789}^o$

So the above equation has 1 real root lying brtween $0$ and $\pi$