The number of values of x with 0 leq x leq 2 | vedclass

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

Question

(B) Given the equation: $\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x$. Group the terms: $(\sin 3x + \sin x) + \sin 2x = (\cos 3x + \cos x)

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Given the equation: $\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x$. Group the terms: $(\sin 3x + \sin x) + \sin 2x = (\cos 3x + \cos x) + \cos 2x$. Using sum-to-product formulas: $2 \sin 2x \cos x + \sin 2x = 2 \cos 2x \cos x + \cos 2x$. Factor out common terms: $\sin 2x (2 \cos x + 1) = \cos 2x (2 \cos x + 1)$. Rearrange: $(2 \cos x + 1)(\sin 2x - \cos 2x) = 0$. Case $1$: $2 \cos x + 1 = 0 \implies \cos x = -1/2$. In the interval $[0, 2\pi]$,$x = 2\pi/3, 4\pi/3$. Case $2$: $\sin 2x - \cos 2x = 0 \implies 	an 2x = 1$. For $0 \leq x \leq 2\pi$,$0 \leq 2x \leq 4\pi$. $2x = \pi/4, 5\pi/4, 9\pi/4, 13\pi/4$. $x = \pi/8, 5\pi/8, 9\pi/8, 13\pi/8$. Total values of $x$ are $2 + 4 = 6$.

The number of values of $x$ with $0 \leq x \leq 2 \pi$ satisfying the equation $\sin x + \sin 2x + \sin 3x = \cos x + \cos 2x + \cos 3x$ is

Similar Questions

If $4\sin^4 x + \cos^4 x = 1$,then $x =$

The solution of the equation $\sec \theta - \text{cosec} \theta = \frac{4}{3}$ is

The number of values of $x$ satisfying $2\sin^2(2x) = 2\cos^2(8x) + \cos(10x)$ in the interval $x \in \left[ -\frac{\pi}{4}, \frac{\pi}{4} \right]$ is:

The general solution of $\sin x - \cos x = \sqrt{2}$,for any integer $n$ is

If $\sin 6 \theta + \sin 4 \theta + \sin 2 \theta = 0$,then the general value of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module