Question

Solve $\sin x+\sqrt{3}\cos x\ge 1$.

Answer 1

Proceeding as in (a), we get
$\cos (x-\frac{\pi }{6})\ge \frac{1}{2}=\cos \frac{\pi }{3}$
$\therefore -\pi /3\le x-\left(\pi /6\right)\le \pi /3$
or $-\pi /6\le x\le \pi /2$.