If the general solution of sin x + 3 sin 3x + | vedclass

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

Question

(D) Given equation: $\sin x + 3 \sin 3x + \sin 5x = 0$ Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$:

Vedclass · Accepted Answer

$\{-1, -\frac{1}{2}, \frac{1}{2}, 1\}$

Vedclass · Answer

(D) Given equation: $\sin x + 3 \sin 3x + \sin 5x = 0$ Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$: $(\sin 5x + \sin x) + 3 \sin 3x = 0$ $2 \sin 3x \cos 2x + 3 \sin 3x = 0$ $\sin 3x (2 \cos 2x + 3) = 0$ Case $1$: $\sin 3x = 0$ $\Rightarrow 3x = n\pi$ $\Rightarrow x = \frac{n\pi}{3}$ Case $2$: $2 \cos 2x + 3 = 0 \Rightarrow \cos 2x = -\frac{3}{2}$ Since the range of $\cos 2x$ is $[-1, 1]$,$\cos 2x = -\frac{3}{2}$ has no real solution. Thus,$x = \frac{n\pi}{3}$. For $x = \frac{n\pi}{3}$,the possible values of $\cos x$ are $\cos(0) = 1$,$\cos(\frac{\pi}{3}) = \frac{1}{2}$,$\cos(\frac{2\pi}{3}) = -\frac{1}{2}$,$\cos(\pi) = -1$,$\cos(\frac{4\pi}{3}) = -\frac{1}{2}$,$\cos(\frac{5\pi}{3}) = \frac{1}{2}$. Therefore,the set of all values of $\cos y$ is $\{-1, -\frac{1}{2}, \frac{1}{2}, 1\}$.

If the general solution of $\sin x + 3 \sin 3x + \sin 5x = 0$ is $x = y$,then the set of all values of $\cos y$ is

Similar Questions

The general solution satisfying both the equations $\sin x = -\frac{3}{5}$ and $\cos x = -\frac{4}{5}$ is

The number of solutions of the equation $2 \theta - \cos^{2} \theta + \sqrt{2} = 0$ is equal to

Find the principal solutions of the equation $\tan x = -\frac{1}{\sqrt{3}}.$

The solution set of the equation $\tan (\pi \tan x) = \cot (\pi \cot x)$ for $x \in (0, \frac{\pi}{2})$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module