Solution of equation {cos ^2}x - 2cos x = 4sin x

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Trigonometrical Equations

hard

The solution of the equation ${\cos ^2}x - 2\cos x = $ $4\sin x - \sin 2x,$ $\,(0 \le x \le \pi )$ is

$\pi - {\cot ^{ - 1}}\left( {\frac{1}{2}} \right)$

$\pi - {\tan ^{ - 1}}(2)$

$\pi + {\tan ^{ - 1}}\left( { - \frac{1}{2}} \right)$

None of these

Solution

(c) Given equation is ${\cos ^2}x – 2\cos x = 4\sin x – \sin 2x$
==> ${\cos ^2}x – 2\cos x = 4\sin x – 2\sin x\cos x$
==> $\cos x(\cos x – 2) = 2\sin x(2 – \cos x)$
==> $(\cos x – 2)(\cos x + 2\sin x) = 0$
==> $\cos x + 2\sin x = 0$,
==> $\tan x = – \frac{1}{2}$ ==> $x = n\pi + {\tan ^{ – 1}}( – 1/2),\,\,n \in I$
As $0 \le x \le \pi ,$ therefore, $x = \pi + {\tan ^{ – 1}}( – 1/2).$

Standard 11

Mathematics

Let $S={\theta \in\left(0, \frac{\pi}{2}\right): \sum_{m=1}^{9}}$

$\sec \left(\theta+(m-1) \frac{\pi}{6}\right) \sec \left(\theta+\frac{m \pi}{6}\right)=-\frac{8}{\sqrt{3}}$ Then.

hard

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