$\begin{align}  & (2+\sqrt{3})\sin 3x-\cos 3x=1 \\ & \text{Divide through by }(2+\sqrt{3}) \\ & \sin 3x-\frac{\cos 3x}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}} \\ & \sin 3x-(2-\sqrt{3})\cos 3x=2-\sqrt{3} \\ & \text{Note:}\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=2-\sqrt{3} \\ & \text{Expressing }\sin 3x-(2-\sqrt{3})\cos 3x\text{ in the form } \\ & R\sin (3x-\alpha ) \\ & \sin 3x-(2-\sqrt{3})\cos 3x=R\sin (3x-\alpha ) \\ & \sin 3x-(2-\sqrt{3})\cos 3x=R\sin 3x\cos \alpha -R\cos 3x\sin \alpha  \\ & \text{Comparing identities} \\ & \sin 3x=R\sin 3x\cos \alpha  \\ & R\cos \alpha =1----(i) \\ & -(2-\sqrt{3})\cos 3x=-R\cos 3x\sin \alpha  \\ & R\sin \alpha =2-\sqrt{3} \\ & \text{Sqauring both sides (i) and (ii) adding them } \\ & \text{together we obtain} \\ & {{R}^{2}}{{\cos }^{2}}\alpha +{{R}^{2}}{{\sin }^{2}}\alpha ={{1}^{2}}+{{(2-\sqrt{3})}^{2}} \\ & {{R}^{2}}({{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha )=8-4\sqrt{3} \\ & R=\sqrt{8-4\sqrt{3}} \\ & \text{Let the square root be }\sqrt{a}-\sqrt{b} \\ & \sqrt{8-4\sqrt{3}}=\sqrt{a}-\sqrt{b} \\ & \text{sqaure both sides} \\ & 8-4\sqrt{3}=a+b-2\sqrt{ab} \\ & a+b=8---(iii) \\ & 4\sqrt{3}=2\sqrt{ab} \\ & ab=12---(iv) \\ & \text{From (iii) }a=8-b \\ & \text{Substitute }8-b\text{ for }a\text{ in (iv)} \\ & (8-b)b=12 \\ & {{b}^{2}}-8b+12=0 \\ & (b-2)(b-6)=0 \\ & b=2\text{ or }b=6 \\ & \text{when }b=2,\text{ }a=6 \\ & \text{when }b=6,\text{ }a=2 \\ & \text{The positive sqaure root will be considered} \\ & \text{which will }R=\sqrt{6}-\sqrt{2} \\ & \text{Divide (ii) by (i)} \\ & \frac{R\sin \alpha }{R\cos \alpha }=\frac{2-\sqrt{3}}{1} \\ & \tan \alpha =2-\sqrt{3} \\ & \alpha ={{\tan }^{-1}}(2-\sqrt{3})={{15}^{\circ }} \\ & \sin 3x-(2-\sqrt{3})\cos 3x=(\sqrt{6}-\sqrt{2})\sin (3x-{{15}^{\circ }}) \\\end{align}$