The solution of the equation cos^2 x - 2cos x | vedclass

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

Question

(C) Given equation: $\cos^2 x - 2\cos x = 4\sin x - \sin 2x$Using $\sin 2x = 2\sin x \cos x$,we get:$\cos^2 x - 2\cos x = 4\sin x - 2\sin x \cos x$$\c

Vedclass · Accepted Answer

$\pi + 	an^{-1}\left(-\frac{1}{2}ight)$

Vedclass · Answer

(C) Given equation: $\cos^2 x - 2\cos x = 4\sin x - \sin 2x$Using $\sin 2x = 2\sin x \cos x$,we get:$\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(\cos x - 2) + 2\sin x(\cos x - 2) = 0$$(\cos x - 2)(\cos x + 2\sin x) = 0$Since $\cos x - 2 
eq 0$ for any real $x$,we must have $\cos x + 2\sin x = 0$.$	an x = -\frac{1}{2}$Since $0 \le x \le \pi$,and $	an x$ is negative in the second quadrant,the solution is $x = \pi + 	an^{-1}\left(-\frac{1}{2}ight)$.

The solution of the equation $\cos^2 x - 2\cos x = 4\sin x - \sin 2x$ for $0 \le x \le \pi$ is

Similar Questions

The general solution of $\tan 3x = 1$ is

The number of principal solutions of the equation $\tan 3x - \tan 2x - \tan x = 0$ is:

The number of solutions of the equation $\tan x + \sec x = 2 \cos x$ for $x \in [0, \pi]$ is

If $\theta \in [-2 \pi, 2 \pi]$,then the number of solutions of $2 \sqrt{2} \cos^2 \theta + (2 - \sqrt{6}) \cos \theta - \sqrt{3} = 0$ is equal to:

If $5\cos 2\theta + 2\cos^2\frac{\theta}{2} + 1 = 0$ and $-\pi < \theta < \pi$,then $\theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module