• question_answer 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, $2\,{{\cot }^{2}}\theta -\frac{5}{\sin \theta }+4=0$, then $\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{{{\cos }^{2}}\,3\theta \,d\theta },$ is equal to [JEE MAIN Held on 07-01-2020 Evening] A) $\frac{\pi }{3}+\frac{1}{6}$           B) $\frac{\pi }{3}$ C) $\frac{2\pi }{3}$                    D) $\frac{\pi }{9}$

[b]  $\because \,\,\,\,2{{\cot }^{2}}\theta -\frac{5}{\sin \,\theta }+4=0$ $2+2\,\text{cose}{{\text{c}}^{2}}\,\theta -5\,\text{cosec}\,\theta =0$ $2\,\text{cose}{{\text{c}}^{2}}\,\theta -4\,\text{cosec}\,\theta -\,\text{cosec}\theta +2=0$ $\therefore \,\,\,(2\,\text{cosec}\,\,\theta -1)\,(\,\text{cosec}\,\theta -2)\,=0$ $\therefore \,\,\text{cosec}\,\theta =\frac{1}{2}$ or 2. $\therefore \,\,\sin \,\theta =2$ or $\frac{1}{2}.$       $\because \,\,\theta \,\in \,(0,2\pi )$ $\therefore \,\,\,\,{{\theta }_{1}}=\frac{\pi }{6}$ and ${{\theta }_{2}}=\frac{5\pi }{6},$                      $\because \,{{\theta }_{1}}<{{\theta }_{2}}$ $\therefore \,\,I=\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}}{{{\cos }^{2}}\,3\theta \,d\theta =\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}}{\frac{1+\cos \,6\theta }{2}}}\,d\theta$ $=\frac{1}{2}\left[ \theta +\frac{\sin \,6\theta }{6} \right]_{\frac{\pi }{6}}^{\frac{5\pi }{6}}=\frac{\pi }{3}$