The general solution of sin x + sin 5x = sin | vedclass

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

Question

(C) Given equation: $\sin x + \sin 5x = \sin 2x + \sin 4x$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2}

Vedclass · Accepted Answer

$n\pi / 3$

Vedclass · Answer

(C) Given equation: $\sin x + \sin 5x = \sin 2x + \sin 4x$Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:$2 \sin(3x) \cos(-2x) = 2 \sin(3x) \cos(-x)$Since $\cos(-	heta) = \cos 	heta$,we have:$2 \sin(3x) \cos(2x) = 2 \sin(3x) \cos(x)$$2 \sin(3x) [\cos(2x) - \cos(x)] = 0$Case $1$: $\sin(3x) = 0 \implies 3x = n\pi \implies x = \frac{n\pi}{3}$Case $2$: $\cos(2x) - \cos(x) = 0 \implies \cos(2x) = \cos(x)$$2x = 2n\pi \pm x$For $+$,$x = 2n\pi$. For $-$,$3x = 2n\pi \implies x = \frac{2n\pi}{3}$Since $\frac{n\pi}{3}$ covers all solutions including $\frac{2n\pi}{3}$ and $2n\pi$,the general solution is $x = \frac{n\pi}{3}$.

The general solution of $\sin x + \sin 5x = \sin 2x + \sin 4x$ is:

Similar Questions

If $\sin \left(5 x+\frac{\pi}{4}\right)=0$,then $x$ is equal to

If $\tan (\cot x) = \cot (\tan x),$ then $\sin 2x =$

The number of values of $x$ in the interval $[0, 5\pi]$ satisfying the equation $3 \sin^2 x - 7 \sin x + 2 = 0$ is

If $\sqrt{3} \cos \theta + \sin \theta = \sqrt{2}$,then the most general value of $\theta$ is

If the equation $\sin 3\theta - \cos^2 \theta = \frac{1}{4}$ and $\theta \in [0, \pi]$, then the number of solutions is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module