$\begin{align}  & \sin x-\sqrt{3}\cos x=R\sin (x-\alpha ) \\ & \sin x-\sqrt{3}\cos x=R\sin x\cos \alpha -R\cos x\sin \alpha  \\ & \text{Comparing identities} \\ & \sin x=R\sin x\cos \alpha  \\ & R\cos \alpha =1----(i) \\ & \text{Square both sides } \\ & {{R}^{2}}{{\cos }^{2}}\alpha =1----(ii) \\ & -\sqrt{3}\cos x=-R\cos x\sin \alpha  \\ & R\sin \alpha =\sqrt{3}----(iii) \\ & \text{Square both sides of (iii)} \\ & {{R}^{2}}{{\sin }^{2}}\alpha =3----(iv) \\ & \text{Add }equation\text{ }(ii)\text{ and (iv) together} \\ & {{R}^{2}}{{\sin }^{2}}\alpha +{{R}^{2}}{{\cos }^{2}}\alpha =1+3 \\ & {{R}^{2}}({{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha )=4 \\ & {{R}^{2}}=4 \\ & R=\pm 2\text{ We need the positive value of }R \\ & R=2 \\ & \text{Divide equation  (iii) by (i)} \\ & \frac{R\sin \alpha }{R\cos \alpha }=\frac{\sqrt{3}}{1} \\ & \tan \alpha =\sqrt{3} \\ & \alpha ={{60}^{\circ }} \\ & \sin x-\sqrt{3}\cos x=2\sin (x-{{60}^{\circ }}) \\ & ii)\text{ }\sqrt{3}\sin x-3\cos x=3 \\ & \text{     Divide through by }\sqrt{3} \\ & \sin x-\frac{3}{\sqrt{3}}\cos x=\frac{3}{\sqrt{3}} \\ & \sin x-\sqrt{3}\cos x=\sqrt{3} \\ & \sin x-\sqrt{3}\cos x=2\sin (x-{{60}^{\circ }})=\sqrt{3} \\ & 2\sin (x-{{60}^{\circ }})=\sqrt{3} \\ & \sin (x-{{60}^{\circ }})=\frac{\sqrt{3}}{2} \\ & (x-{{60}^{\circ }})={{\sin }^{-1}}\frac{\sqrt{3}}{2} \\ & x-{{60}^{\circ }}={{60}^{\circ }} \\ & \text{The general solution } \\ & x-{{60}^{\circ }}=n{{180}^{\circ }}+{{(-1)}^{n}}{{60}^{\circ }} \\ & \text{When }n=0 \\ & x-{{60}^{\circ }}={{60}^{\circ }},\text{  }x={{120}^{\circ }} \\ & \text{when }n=1 \\ & x-{{60}^{\circ }}={{120}^{\circ }},\text{  }x={{180}^{\circ }} \\ & x={{120}^{\circ }},x={{180}^{\circ }} \\\end{align}$