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

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

Question

(C) Given equation: $\sin x + \sin 5x + 3 \sin 3x = 0$ Using the formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get: $2 \sin

Vedclass · Accepted Answer

$\{\frac{\sqrt{3}}{2}, 0, -\frac{\sqrt{3}}{2}\}$

Vedclass · Answer

(C) Given equation: $\sin x + \sin 5x + 3 \sin 3x = 0$ Using the formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get: $2 \sin 3x \cos 2x + 3 \sin 3x = 0$ $\sin 3x (2 \cos 2x + 3) = 0$ Since $2 \cos 2x + 3 = 0$ implies $\cos 2x = -\frac{3}{2}$,which is impossible as $-1 \le \cos 2x \le 1$,we must have $\sin 3x = 0$. Thus,$3x = n\pi$,or $x = \frac{n\pi}{3}$ for $n \in \mathbb{Z}$. The set of values for $\sin x$ is $\{\sin(0), \sin(\frac{\pi}{3}), \sin(\frac{2\pi}{3}), \sin(\pi), \sin(\frac{4\pi}{3}), \sin(\frac{5\pi}{3})\}$. Evaluating these: $\{0, \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 0, -\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}\}$. The set of distinct values is $\{0, \frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}\}$.

If the general solution set of $\sin x + 3 \sin 3x + \sin 5x = 0$ is $S$,then $\{\sin \alpha \mid \alpha \in S\} = $

Similar Questions

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

The general solution of $2 \sqrt{3} \cos^2 \theta = \sin \theta$ is

The number of solutions of the equation $\sin A - 5 \sin 2A + \sin 3A = \cos A - 5 \cos 2A + \cos 3A$ in the interval $(0, \pi)$ is:

The smallest positive value of $x$ in degrees satisfying the equation $\tan(x+100^{\circ}) = \tan(x+50^{\circ}) \tan(x) \tan(x-50^{\circ})$ is (in $^{\circ}$)

The general solution of $\sin^{2} x \cdot \sec x = \tan x - \sin x + 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module