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

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Question

(a) We have $\cos 3x + \sin {\rm{ }}\left( {2x - \frac{{7\pi }}{6}} \right)\, = - 2$

$ \Rightarrow $ $1 + \cos 3x + 1 + \sin \left( {2x - \frac{{7\

Vedclass · Accepted Answer

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

Vedclass · Answer

(a) We have $\cos 3x + \sin {\rm{ }}\left( {2x - \frac{{7\pi }}{6}} \right)\, = - 2$

$ \Rightarrow $ $1 + \cos 3x + 1 + \sin \left( {2x - \frac{{7\pi }}{6}} \right) = 0$

$ \Rightarrow $ $(1 + \cos 3x) + 1 - \cos \left( {2x - \frac{{2\pi }}{3}} \right) = 0$

$ \Rightarrow $ $2{\cos ^2}\frac{{3x}}{2} + 2{\sin ^2}\left( {x - \frac{\pi }{3}} \right) = 0$

$ \Rightarrow $ $\cos \frac{{3x}}{2} = 0$ and $\sin \left( {x - \frac{\pi }{3}} \right) = 0$

$ \Rightarrow $$\frac{{3x}}{2} = \frac{\pi }{2},\,\frac{{3\pi }}{2},\,.....$

and $x - \frac{\pi }{3}$=0, $\pi ,2\pi ..... \Rightarrow x = \frac{\pi }{3}$

Therefore, the general solution of $\cos \frac{{3x}}{2} = 0$

and $\sin \left( {x - \frac{\pi }{3}} \right) = 0$ is $x = 2k\pi + \frac{\pi }{3} = \frac{\pi }{3}(6k + 1),$ where $k \in Z$.

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

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