Question

Solve : $\sin \theta +\sqrt{3}\cos \theta \ge 1$, $-\pi <\theta \le \pi$

• A. $-\frac{\pi }{3}\le \theta \le \frac{\pi }{2}$
• B. $-\frac{\pi }{6}\le \theta \le \frac{\pi }{4}$
• C. $-\frac{\pi }{8}\le \theta \le \frac{\pi }{4}$
• D. $-\frac{\pi }{6}\le \theta \le \frac{\pi }{2}$

Answer 1

The correct option is D $-\frac{\pi }{6}\le \theta \le \frac{\pi }{2}$

The given inequation is
$\sin \theta +\sqrt{3}\cos \theta \ge 1$, $-\pi <\theta \le \pi$

$\Rightarrow$ $\frac{1}{2}\sin \theta +\frac{\sqrt{3}}{2}\cos \theta \ge \frac{1}{2}$

$\Rightarrow$ $\sin \frac{\pi }{6}\sin \theta +\cos \frac{\pi }{6}\cos \theta \ge \frac{1}{2}$

$\Rightarrow$ $\cos (\theta -\frac{\pi }{6})\ge \frac{1}{2}=\cos \frac{\pi }{3}$

$\Rightarrow$ $-\frac{\pi }{3}\le \theta -\frac{\pi }{6}\le \frac{\pi }{3}$

$\Rightarrow$ $-\frac{\pi }{6}\le \theta \le \frac{\pi }{2}$