The solutions of sin x + sin 5x = sin 3x in t | vedclass

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

Question

(B) Given equation: $\sin x + \sin 5x = \sin 3x$ Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$: $2 \s

Vedclass · Accepted Answer

$\frac{\pi}{6}, \frac{\pi}{3}$

Vedclass · Answer

(B) Given equation: $\sin x + \sin 5x = \sin 3x$ Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$: $2 \sin(\frac{x+5x}{2}) \cos(\frac{x-5x}{2}) = \sin 3x$ $2 \sin 3x \cos(-2x) = \sin 3x$ Since $\cos(-2x) = \cos 2x$,we have: $2 \sin 3x \cos 2x - \sin 3x = 0$ $\sin 3x (2 \cos 2x - 1) = 0$ Case $1$: $\sin 3x = 0 \implies 3x = n\pi \implies x = \frac{n\pi}{3}$. For $x \in (0, \frac{\pi}{2})$,$x = \frac{\pi}{3}$ is a solution. Case $2$: $2 \cos 2x - 1 = 0 \implies \cos 2x = \frac{1}{2} = \cos \frac{\pi}{3}$. $2x = 2n\pi \pm \frac{\pi}{3} \implies x = n\pi \pm \frac{\pi}{6}$. For $x \in (0, \frac{\pi}{2})$,$x = \frac{\pi}{6}$ is a solution. Thus,the solutions are $x = \frac{\pi}{6}, \frac{\pi}{3}$.

The solutions of $\sin x + \sin 5x = \sin 3x$ in the interval $(0, \frac{\pi}{2})$ are

Similar Questions

When $a$ is irrational,the number of solutions satisfying the equation $1+\sin^2(ax)=\cos(x)$ is

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

The solution set of the equation $\tan x + \sec x = 2 \cos x$ in the interval $[0, 2 \pi]$ is

The value of $\cos y \cos (\frac{\pi}{2} - x) - \cos (\frac{\pi}{2} - y) \cos x + \sin y \cos (\frac{\pi}{2} - x) + \cos x \sin (\frac{\pi}{2} - y)$ is zero,if

If the solution of the equation $\log _{\cos x} \cot x+4 \log _{\sin x} \tan x=1$,where $x \in \left(0, \frac{\pi}{2}\right)$,is $\sin ^{-1}\left(\frac{\alpha+\sqrt{\beta}}{2}\right)$,where $\alpha, \beta$ are integers,then $\alpha+\beta$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module