The number of solutions of the equation sin 2 | vedclass

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

Question

(A) Given equation: $\sin 2x - 2 \cos x + 4 \sin x = 4$Using $\sin 2x = 2 \sin x \cos x$,we get:$2 \sin x \cos x - 2 \cos x + 4 \sin x - 4 = 0$Factor

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Given equation: $\sin 2x - 2 \cos x + 4 \sin x = 4$Using $\sin 2x = 2 \sin x \cos x$,we get:$2 \sin x \cos x - 2 \cos x + 4 \sin x - 4 = 0$Factor by grouping:$2 \cos x (\sin x - 1) + 4 (\sin x - 1) = 0$$(2 \cos x + 4)(\sin x - 1) = 0$$2(\cos x + 2)(\sin x - 1) = 0$Since $\cos x + 2 
eq 0$ for any real $x$ (as $-1 \le \cos x \le 1$),we must have:$\sin x = 1$In the interval $[0, 5\pi]$,the values of $x$ for which $\sin x = 1$ are:$x = \frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}$Thus,there are $3$ solutions.

The number of solutions of the equation $\sin 2x - 2 \cos x + 4 \sin x = 4$ in the interval $[0, 5\pi]$ is

Similar Questions

If $\sin 2\theta = \cos \theta$ and $0 < \theta < \pi$,then the possible values of $\theta$ are:

If $\cos \theta \neq 0$,and $\sec \theta - 1 = (\sqrt{2} - 1) \tan \theta$,then $\theta =$

If $4(\sin 2x \sin 4x + \sin^2 x) = 3$,then $x =$

The number of solutions of $\cos 2 \theta = \sin \theta$ in the interval $(0, 2 \pi)$ is:

If $\cos 2x = (\sqrt{2}+1)(\cos x - \frac{1}{\sqrt{2}})$ and $\cos x \neq \frac{1}{\sqrt{2}}$,then $x \in$

Vedclass Test Series

Exam Paper Generator

Online Exam Module