If cos⁡3x+sin⁡2x−7π6=−2 then x=

A
$\frac{π}{3} (6 k + 1), k \in Z$
B
$\frac{π}{3} (6 k - 1), k \in Z$
C
$\frac{π}{3} (2 k + 1), k \in Z$
D
none of these

Solution:

We have

$\begin{array}{l} \cos 3 x + \sin (2 x - \frac{7 π}{6}) = - 2 \\ \Rightarrow 1 + \cos 3 x + 1 + \sin (2 x - \frac{7 π}{6}) = 0 \\ \Rightarrow (1 + \cos 3 x) + 1 - \cos (2 x - \frac{2 π}{3}) = 0 \\ \Rightarrow 2 \cos^{2} \frac{3 x}{2} + 2 \sin^{2} (x - \frac{π}{3}) = 0 \\ \Rightarrow \cos \frac{3 x}{2} = 0 and \sin (x - \frac{π}{3}) = 0 \\ \Rightarrow \frac{3 x}{2} = \frac{π}{2}, \frac{3 π}{2}, \dots and x - \frac{π}{3} = 0, π, 2 π \dots \\ \Rightarrow x = \frac{π}{3} \end{array}$

Therefore, the general solution is given by

$x = 2 k π + \frac{π}{3} = \frac{π}{3} (6 k + 1), where k \in Z .$

If $\cos 3 x + \sin (2 x - \frac{7 π}{6}) = - 2$ then $x =$

Solution:

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