Question is as below:

$f\left(x\right) = \mathrm{cosx}-\sqrt{3}\mathrm{sinx}=2\left(\frac{1}{2}\mathrm{cosx}-\frac{\sqrt{3}}{2}\mathrm{sinx}\right)$$=2\cos \left(\frac{\pi }{3}+x\right) \mathrm{using} \mathrm{formula} \cos (A+B)$$-2⩽f\left(x\right)⩽2$$-2⩽2\cos \left(\frac{\pi }{3}+x\right)⩽2$$-1⩽\cos \left(\frac{\pi }{3}+x\right)⩽1$$\mathrm{hence} 0⩽\left(\frac{\pi }{3}+x\right)⩽\pi$$-\frac{\pi }{3}⩽x⩽\pi -\frac{\pi }{3}$$-\frac{\pi }{3}⩽x⩽\frac{2\pi }{3} \mathrm{so} \mathrm{domain} =D=\left[-\frac{\pi }{3},\frac{2\pi }{3} \right]$">