The general solution of sin x - 3sin 2x + sin | vedclass

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

Question

(B) Given equation: $\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x$Rearranging terms: $(\sin 3x + \sin x) - 3\sin 2x = (\cos 3x + \cos x)

Vedclass · Accepted Answer

$\frac{n\pi}{2} + \frac{\pi}{8}$

Vedclass · Answer

(B) Given equation: $\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x$Rearranging terms: $(\sin 3x + \sin x) - 3\sin 2x = (\cos 3x + \cos x) - 3\cos 2x$Using sum-to-product formulas: $2\sin 2x \cos x - 3\sin 2x = 2\cos 2x \cos x - 3\cos 2x$$\sin 2x(2\cos x - 3) = \cos 2x(2\cos x - 3)$$(\sin 2x - \cos 2x)(2\cos x - 3) = 0$Since $2\cos x - 3 
eq 0$ (as $\cos x$ cannot be $1.5$),we have $\sin 2x = \cos 2x$$	an 2x = 1$$2x = n\pi + \frac{\pi}{4}$$x = \frac{n\pi}{2} + \frac{\pi}{8}$

The general solution of $\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x$ is

Similar Questions

The number of solutions of $\sin^{7} x + \cos^{7} x = 1$ for $x \in [0, 4\pi]$ is equal to:

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:

The number of solutions of $8 \cos x = x$ will be:

If $12 \cot^2 \theta - 31 \csc \theta + 32 = 0$,then the value of $\sin \theta$ is

If $\sin \theta = \sqrt{3} \cos \theta$ and $-\pi < \theta < 0$,then $\theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module