The sum of the solutions in x in (0, 4pi) of | vedclass

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

Question

(C) Using the identity $\sin 	heta \sin(\frac{\pi}{3} - 	heta) \sin(\frac{\pi}{3} + 	heta) = \frac{1}{4} \sin(3	heta)$,we let $	heta = \frac{x}{3

Vedclass · Accepted Answer

$3\pi$

Vedclass · Answer

(C) Using the identity $\sin 	heta \sin(\frac{\pi}{3} - 	heta) \sin(\frac{\pi}{3} + 	heta) = \frac{1}{4} \sin(3	heta)$,we let $	heta = \frac{x}{3}$.Then the equation becomes $4 \sin \frac{x}{3} \sin(\frac{\pi}{3} + \frac{x}{3}) \sin(\frac{2\pi}{3} + \frac{x}{3}) = 1$.Note that $\sin(\frac{2\pi}{3} + \frac{x}{3}) = \sin(\pi - (\frac{\pi}{3} - \frac{x}{3})) = \sin(\frac{\pi}{3} - \frac{x}{3})$.Thus,the expression is $4 \sin \frac{x}{3} \sin(\frac{\pi}{3} + \frac{x}{3}) \sin(\frac{\pi}{3} - \frac{x}{3}) = 4 	imes \frac{1}{4} \sin(3 	imes \frac{x}{3}) = \sin x$.So,$\sin x = 1$.Given $x \in (0, 4\pi)$,the solutions are $x = \frac{\pi}{2}$ and $x = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2}$.The sum of the solutions is $\frac{\pi}{2} + \frac{5\pi}{2} = \frac{6\pi}{2} = 3\pi$.

The sum of the solutions in $x \in (0, 4\pi)$ of the equation $4\sin \frac{x}{3} \sin \left( \frac{\pi + x}{3} \right) \sin \left( \frac{2\pi + x}{3} \right) = 1$ is

Similar Questions

The number of distinct solutions of the equation $\frac{5}{4} \cos ^2 2x + \cos ^4 x + \sin ^4 x + \cos ^6 x + \sin ^6 x = 2$ in the interval $[0, 2\pi]$ is

The set of solutions of the equation $(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$ is

The total number of solutions of $\sin^4x + \cos^4x = \sin x \cos x$ in the interval $[0, 2\pi]$ is equal to

For $n \in \mathbb{Z}$,the general solution of the trigonometric equation $\sin x - \sqrt{3} \cos x + 4 \sin 2x - 4 \sqrt{3} \cos 2x + \sin 3x - \sqrt{3} \cos 3x = 0$ is

The number of solutions of $\sec x \cos 5x + 1 = 0$ in the interval $[0, 2\pi]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module