Question

If $\sqrt{3}\cos x-\sin x=1$, then $x$ is

• A. $2n\pi \pm \frac{\pi }{4}-\frac{\pi }{6},n\in I$
• B. $2n\pi \pm \frac{\pi }{6}-\frac{\pi }{3},n\in I$
• C. $2n\pi \pm \frac{\pi }{3}-\frac{\pi }{6},n\in I$
• D. $Noneofthese$

Answer 1

The correct option is C $2n\pi \pm \frac{\pi }{3}-\frac{\pi }{6},n\in I$

Given

$\sqrt{3}\cos x-\sin x=1$

Dividing by $2$ on both sides

$⟹\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x=\frac{1}{2}$

$⟹\cos \frac{\pi }{6}\cos x-\sin \frac{\pi }{6}\sin x=\cos \frac{\pi }{3}$

$⟹\cos (\frac{\pi }{6}+x)=\cos \frac{\pi }{3}$

$(\cos (A+B)=\cos A\cos B-\sin A\sin B)$

$⟹\frac{\pi }{6}+x=2n\pi \pm \frac{\pi }{3},n\in I$

$(\cos x=\cos k⟹x=2n\pi \pm k)$

$⟹x=2n\pi \pm \frac{\pi }{3}-\frac{\pi }{6},n\in I$