સમીકરણ 8cos x cdot left( {cos left( {frac{pi | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$8 \cos x\left\{\cos \left(\frac{\pi}{6}+x\right) \cdot \cos \left(\frac{\pi}{6}-x\right)-\frac{1}{2}\right\}=1$

$\Rightarrow 4 \cos x\left\{2 \cos

Vedclass · Accepted Answer

$\frac{{13}}{9}$

Vedclass · Answer

$8 \cos x\left\{\cos \left(\frac{\pi}{6}+xight) \cdot \cos \left(\frac{\pi}{6}-xight)-\frac{1}{2}ight\}=1$

$\Rightarrow 4 \cos x\left\{2 \cos \left(\frac{\pi}{6}+xight) \cos \left(\frac{\pi}{6}-xight)-1ight\}=1$

$\Rightarrow 4 \cos x\left\{\cos 2 x+\cos \frac{\pi}{3}-1ight\}=1$

$\Rightarrow 4 \cos x\left\{\cos 2 x-\frac{1}{2}ight\}=1$

$\Rightarrow 4 \cos x\left\{2 \cos ^{2} x-1-\frac{1}{2}ight\}=1$

$\Rightarrow 8 \cos ^{3} x-6 \cos x=1$

$\Rightarrow 2\left(4 \cos ^{3} x-3 \cos xight)=1$

$ \cos 3 x=\frac{1}{2}$

$\cos 3x = \cos {60^\circ },\cos {30^\circ },\cos {420^\circ }$

$x{	ext{ }} = {20^\circ },{100^\circ },{140^\circ }$

${20^\circ } + {100^\circ } + {140^\circ }{	ext{ }} = k\pi $

$ \Rightarrow {260^\circ } = k\left( {{{180}^\circ }} ight)$

$ \Rightarrow k = \frac{{{{260}^\circ }}}{{{{180}^\circ }}} = \frac{{13}}{9}$

Similar Questions

જો $\sin {\rm{ }}\left( {\frac{\pi }{4}\cot \theta } \right) = \cos {\rm{ }}\left( {\frac{\pi }{4}\tan \theta } \right)\,\,,$ તો $\theta = $

જો $2\sin \theta + \tan \theta = 0$ નું સમાધાન કરે તેવા $\theta $ નો વ્યાપક ઉકેલ મેળવો.

સમીકરણ $4 \sin ^2 x-4 \cos ^3 x+9-4 \cos x=0 ; x \in[-2 \pi, 2 \pi]$ નાં ઉકેલોની સંખ્યા __________છે.

સમીકરણ $ln(1 + sin^2x) = 1 -ln(5 + x^2)$ ના ઉકેલોની સંખ્યા મેળવો

જો $L=\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$ અને $M=\cos ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right),$ હોય તો