# Prove the following

Question:

If $\theta_{1}$ and $\theta_{2}$ be respectively the smallest and the largest values of $\theta$ in $(0,2 \pi)-\{\pi\}$ which satisfy the equation,

1. (1) $\frac{\pi}{3}$

2. (2) $\frac{2 \pi}{3}$

3. (3) $\frac{\pi}{3}+\frac{1}{6}$

4. (4) $\frac{\pi}{9}$

Correct Option: 1

Solution:

$2 \cot ^{2} \theta-\frac{5}{\sin \theta}+4=0$

$\frac{2 \cos ^{2} \theta}{\sin ^{2} \theta}-\frac{5}{\sin \theta}+4=0$

$\Rightarrow \quad 2 \cos ^{2} \theta-5 \sin \theta+4 \sin ^{2} \theta=0, \sin \theta \neq 0$

$\Rightarrow 2 \sin ^{2} \theta-5 \sin \theta+2=0$

$\Rightarrow(2 \sin \theta-1)(\sin \theta-2)=0$

$\therefore \quad \sin \theta=\frac{1}{2} \quad \Rightarrow \quad \theta=\frac{\pi}{6}, \frac{5 \pi}{6}$

$\therefore \quad \int_{\pi / 6}^{5 \pi / 6} \cos ^{2} 3 \theta d \theta=\int_{\pi / 6}^{5 \pi / 6} \frac{1+\cos 6 \theta}{2} d \theta$

$=\frac{1}{2}\left[\theta+\frac{\sin 6 \theta}{6}\right]_{\pi / 6}^{5 \pi / 6}=\frac{1}{2}\left[\frac{5 \pi}{6}-\frac{\pi}{6}+\frac{1}{6}(0-0)\right]$

$=\frac{1}{2} \cdot \frac{4 \pi}{6}=\frac{\pi}{3}$