Solve this:

Question No. 18 
What is the general solution of :

$$\begin{gathered} \sin 2\mathrm{x}-2 \sin \mathrm{x}-2 {\sin }^2\frac{\mathrm{x}}{2}=0? \\ \left(\mathrm{A}\right) \mathrm{x}=\mathrm{n\pi } \mathrm{or} \mathrm{n\pi }+{\left(-1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{6}, \mathrm{n}\in \mathrm{Z} \\ \left(\mathrm{B}\right) \mathrm{x}=\mathrm{n\pi } \mathrm{or} \mathrm{n\pi }+{\left(-1\right)}^{\mathrm{n}+1}\frac{\mathrm{\pi }}{6}, \mathrm{n}\in \mathrm{Z} \\ \left(\mathrm{C}\right) \mathrm{x}=2\mathrm{n\pi } \mathrm{or} \mathrm{n\pi }+{\left(-1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{6}, \mathrm{n}\in \mathrm{Z} \\ \left(\mathrm{D}\right) \mathrm{x}=2\mathrm{n\pi } \mathrm{or} \mathrm{n\pi }+{\left(-1\right)}^{\mathrm{n}+1}\frac{\mathrm{\pi }}{6}, \mathrm{n}\in \mathrm{Z} \end{gathered}$$