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 which of the following statements is/are correct $?$

• A. ${\theta }_1+{\theta }_2=\pi$
• B. ${\theta }_1+{\theta }_2=\frac{5\pi }{6}$
• C. $\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\cos }^{2}3\theta d\theta =\frac{\pi }{3}$
• D. $\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\cos }^{2}3\theta d\theta =\pi$

Answer 1

Answer: (A), (C)

$\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\cos }^{2}3\theta d\theta =\frac{\pi }{3}$

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

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

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

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

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

$\Rightarrow 2\sin \theta -1=0$ or $\sin \theta \ne 2$

$\Rightarrow \sin \theta =\frac{1}{2}$

$\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6}$ as $\theta$ lies in $(0,2\pi )-\left\{\pi \right\}$

$\therefore {\theta }_1=\frac{\pi }{6},{\theta }_2=\frac{5\pi }{6}$

$\therefore \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}{\cos }^{2}3\theta =\underset{\pi /6}{\overset{5\pi /6}{\int }}\frac{1+\cos 6\theta }{2}d\theta$

$=\frac{1}{2}{[\theta +\frac{\sin 6\theta }{6}]}_{\pi /6}^{5\pi /6}$

$=\frac{1}{2}[\frac{5\pi }{6}+\frac{\sin 5\pi }{6}-\frac{\pi }{6}-\frac{\sin \pi }{6}]=\frac{1}{2}\left[\frac{2\pi }{3}\right]$

$=\frac{\pi }{3}$