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[JEE 2010, 3]
$\frac{1}{\sin \left(\frac{\pi}{\mathrm{n}}\right)}=\frac{1}{\sin \left(\frac{2 \pi}{\mathrm{n}}\right)}+\frac{1}{\sin \left(\frac{3 \pi}{\mathrm{n}}\right)}$ is [JEE 2011, 4]
$\frac{1}{\sin \frac{\pi}{\mathrm{n}}}=\frac{1}{\sin \frac{2 \pi}{\mathrm{n}}}+\frac{1}{\sin \frac{3 \pi}{\mathrm{n}}}$ $\Rightarrow \frac{1}{\sin \frac{\pi}{\mathrm{n}}}-\frac{1}{\sin \frac{3 \pi}{\mathrm{n}}}=\frac{1}{\sin \frac{2 \pi}{\mathrm{n}}}$ $\Rightarrow \frac{\sin \frac{3 \pi}{n}-\sin \frac{\pi}{n}}{\sin \frac{\pi}{n} \sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}$ $\Rightarrow \frac{2 \cos \frac{2 \pi}{\mathrm{n}} \sin \frac{\pi}{\mathrm{n}}}{\sin \frac{\pi}{\mathrm{n}} \sin \frac{3 \pi}{\mathrm{n}}}=\frac{1}{\sin \frac{2 \pi}{\mathrm{n}}}$ $\Rightarrow 2 \cos \frac{2 \pi}{\mathrm{n}} \sin \frac{2 \pi}{\mathrm{n}}=\sin \frac{3 \pi}{\mathrm{n}}$ $\Rightarrow \sin \frac{4 \pi}{\mathrm{n}}=\sin \frac{3 \pi}{\mathrm{n}} \Rightarrow \frac{4 \pi}{\mathrm{n}}=\mathrm{K} \pi+(-1)^{\mathrm{K}} \frac{3 \pi}{\mathrm{n}}$ If $\mathrm{K}=2 \mathrm{m} \quad \Rightarrow \quad \frac{\pi}{\mathrm{n}}=2 \mathrm{m} \pi$ $\Rightarrow \quad n=\frac{1}{2 m} \quad \Rightarrow n=\frac{1}{2}, \frac{1}{4}, \frac{1}{6} \ldots \ldots$ If $\mathrm{K}=2 \mathrm{m}+1 \Rightarrow \frac{7 \pi}{\mathrm{n}}=(2 \mathrm{m}+1) \pi$ $\Rightarrow \mathrm{n}=\frac{7}{2 \mathrm{m}+1} \quad \Rightarrow \quad \mathrm{n}=7, \frac{7}{3}, \frac{7}{5} \ldots \ldots$ Possible value of n is 7
and $-1<\sin \theta<-\frac{\sqrt{3}}{2} .$ Then $\varphi$ cannot satisfy- (A) $0<\varphi<\frac{\pi}{2}$ (B) $\frac{\pi}{2}<\varphi<\frac{4 \pi}{3}$ (C) $\frac{4 \pi}{3}<\varphi<\frac{3 \pi}{2}$ (D) $\frac{3 \pi}{2}<\varphi<2 \pi$ [JEE 2012, 4M]
(A) infinitely many solutions (B) three solutions (C) one solution (D) no solution [JEE(Advanced)-2014, 3(–1)]
[JEE 2015, 4M, –0M]
(A) $-\frac{7 \pi}{9}$ (B) $-\frac{2 \pi}{9}$ (C) 0 (D) $\frac{5 \pi}{9}$ [JEE(Advanced)-2016]
$\sqrt{3} \sin x+\cos x=2 \cos 2 x$ $\Rightarrow \cos 2 x=\cos \left(x-\frac{\pi}{3}\right)$ $\Rightarrow 2 x=2 n \pi \pm\left(x-\frac{\pi}{3}\right)$ $\quad \quad x=(6 n-1) \frac{\pi}{3}$ or $(6 n+1) \frac{\pi}{9}$ $\Rightarrow x=-\frac{\pi}{3}, \frac{\pi}{9}, \frac{7 \pi}{9}$ and $-\frac{5 \pi}{9}$ in $(-\pi, \pi)$ $\Rightarrow \operatorname{sum}=0$
$=$ [JEE(Advanced)-2018]
