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 $S=\left\{\theta \in[-\pi, \pi]-\left\{\pm \frac{\pi}{2}\right\}: \sin \theta \tan \theta+\tan \theta=\sin 2 \theta\right\} \text {. }$ If $T =\sum_{\theta \in S } \cos 2 \theta$, then $T + n ( S )$ is equal

hard

(JEE MAIN-2022)

For which value of $x$ ; $cosx > sinx,$ where $x\, \in \,\,\left( {\frac{\pi }{2}\,,\,\frac{{3\pi }}{2}} \right)$

normal

The most general value of $\theta $ which will satisfy both the equations $\sin \theta = – \frac{1}{2}$ and $\tan \theta = \frac{1}{{\sqrt 3 }}$ is

easy

The sum of all values of $\theta \in[0,2 \pi]$ satisfying $2 \sin ^2 \theta=\cos 2 \theta$ and $2 \cos ^2 \theta=3 \sin \theta$ is

hard

(JEE MAIN-2025)

General solution of $\tan 5\theta = \cot 2\theta $ is $($ where $n \in Z )$

easy

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

Solution

Similar Questions

Let $S=\left\{\theta \in[-\pi, \pi]-\left\{\pm \frac{\pi}{2}\right\}: \sin \theta \tan \theta+\tan \theta=\sin 2 \theta\right\} \text {. }$ If $T =\sum_{\theta \in S } \cos 2 \theta$, then $T + n ( S )$ is equal

For which value of $x$ ; $cosx > sinx,$ where $x\, \in \,\,\left( {\frac{\pi }{2}\,,\,\frac{{3\pi }}{2}} \right)$

The most general value of $\theta $ which will satisfy both the equations $\sin \theta = – \frac{1}{2}$ and $\tan \theta = \frac{1}{{\sqrt 3 }}$ is

The sum of all values of $\theta \in[0,2 \pi]$ satisfying $2 \sin ^2 \theta=\cos 2 \theta$ and $2 \cos ^2 \theta=3 \sin \theta$ is

General solution of $\tan 5\theta = \cot 2\theta $ is $($ where $n \in Z )$

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