$If (2+\sqrt{3}) \cos x=1-\sin x then x =$

$$\begin{gathered} \left(2+\sqrt{3}\right)\cos \mathrm{x}=1-\sin \mathrm{x} \\ \cot {15}^0 \cos \mathrm{x}=1-\sin \mathrm{x} \\ \cos {15}^0 \cos \mathrm{x}=\sin {15}^0-\sin \mathrm{x} \sin {15}^0 \\ \cos {15}^0 \cos \mathrm{x}+\sin {15}^0 \sin \mathrm{x}=\sin {15}^0 \\ \cos (\mathrm{x}-{15}^0)=\cos {75}^0 \\ \mathrm{x}-{15}^0=2\mathrm{n\pi }\pm {75}^0 \\ \mathrm{x}=2\mathrm{n\pi }+\frac{\mathrm{\pi }}{2} \\ \mathrm{x}=2\mathrm{n\pi }-\frac{\mathrm{\pi }}{3} \end{gathered}$$