If 3 x+ (2 x-(7 π)/(6))=-2 , then x is equal to ( where m ∈ I)

$$\begin{aligned} & 1+\cos 3 x+1-\left[\cos \left(\frac{\pi}{2}+\left(2 x-\frac{7 \pi}{6}\right)\right)\right]=0 \\ & 2 \cos ^2 \frac{3 x}{2}+1-\cos \left(2 x-\frac{2 \pi}{3}\right)=0 \\ & 2 \cos ^2 \frac{3 x}{2}+2 \sin ^2\left(x-\frac{\pi}{3}\right)=0\end{aligned}$$\cos \frac{3 x}{2}=0, \sin \left(x-\frac{\pi}{3}\right)=0\mathrm{x}=\frac{\pi}{3}, \pi$and$\mathrm{x}=\frac{\pi}{3}, \frac{2 \pi}{3}, \frac{7 \pi}{3}, \ldots\therefore \mathrm{x}=\frac{\pi}{3}$is the common value which satisfies both$\therefore x=2 n \pi+\frac{\pi}{3}=(6 n+1) \frac{\pi}{3}$