Question

If ${\theta }_1$ and ${\theta }_2$ be respectively the smallest and the largest values of $\theta$ in $(0,2\pi )-\left\{\pi \right\}$ which satisfy the equation, $2{\cot }^2\theta -\frac{5}{\sin \theta }+4=0$, then $\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\cos }^{2}3\theta d\theta$ is equal to :

• A. $\frac{2\pi }{3}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{3}+\frac{1}{6}$
• D. $\frac{\pi }{9}$

Answer 1

The correct option is B $\frac{\pi }{3}$

$2{\cot }^2\theta -\frac{5}{\sin \theta }+4=0,\theta \in [0,2\pi ]$
$\Rightarrow 2{\text{cosec}}^2\theta -2-5\text{cosec}\theta +4=0$
$\Rightarrow 2{\text{cosec}}^2\theta -5\text{cosec}\theta +2=0$
$\Rightarrow \text{cosec}\theta =2$ or $\frac{1}{2}$
$\Rightarrow \text{cosec}\theta =2$
(As $\text{cosec}\theta =\frac{1}{2}$ is not possible)

As $\theta \in [0,2\pi ]$
${\theta }_1=\frac{\pi }{6},{\theta }_2=\frac{5\pi }{6}$
$\Rightarrow \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\cos }^{2}3\theta d\theta =\underset{\pi /6}{\overset{5\pi /6}{\int }}\frac{(1+\cos 6\theta )}{2}d\theta$
$=\frac{1}{2}(\frac{5\pi }{6}-\frac{\pi }{6})+{\left[\frac{\sin 6\theta }{12}\right]}_{\pi /6}^{5\pi /6}$
$=\frac{\pi }{3}$