If cos 3x + sin left( 2x - frac{7pi}{6} right | vedclass

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(A) Given the equation $\cos 3x + \sin \left( 2x - \frac{7\pi}{6} ight) = -2$.Since the range of $\cos 	heta$ and $\sin 	heta$ is $[-1, 1]$,the su

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$\frac{\pi}{3}(6k + 1)$

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(A) Given the equation $\cos 3x + \sin \left( 2x - \frac{7\pi}{6} ight) = -2$.Since the range of $\cos 	heta$ and $\sin 	heta$ is $[-1, 1]$,the sum can be $-2$ only if $\cos 3x = -1$ and $\sin \left( 2x - \frac{7\pi}{6} ight) = -1$.For $\cos 3x = -1$,$3x = (2k + 1)\pi \Rightarrow x = \frac{(2k + 1)\pi}{3}$.For $\sin \left( 2x - \frac{7\pi}{6} ight) = -1$,$2x - \frac{7\pi}{6} = 2n\pi - \frac{\pi}{2}$ $\Rightarrow 2x = 2n\pi + \frac{2\pi}{3}$ $\Rightarrow x = n\pi + \frac{\pi}{3}$.Equating the values of $x$,we find that $x = \frac{\pi}{3}(6k + 1)$ satisfies both conditions.

If $\cos 3x + \sin \left( 2x - \frac{7\pi}{6} \right) = -2$,then $x = $ (where $k \in Z$)

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