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

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(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)

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$6$

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(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

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