Question

Solution set of $\theta$ for which $\sin \theta +\sqrt{3}\cos \theta \ge 1;-\pi <\theta \le \pi$

• A. $\left(\begin{array}{c}-\frac{\pi }{6},\frac{\pi }{6}\end{array}\right)$
• B. $\left[\begin{array}{c}-\frac{\pi }{6},\frac{\pi }{2}\end{array}\right]$
• C. $\left[\begin{array}{c}-\frac{\pi }{3},\frac{\pi }{3}\end{array}\right]$
• D. $\left[\begin{array}{c}-\frac{\pi }{6},\frac{\pi }{2}\end{array}\right]$

Answer 1

Answer: (B)

$\left[\begin{array}{c}-\frac{\pi }{6},\frac{\pi }{2}\end{array}\right]$

$\sin \theta +\sqrt{3}\cos \theta \ge 1,-\pi <\theta \le \pi$
or $\frac{1}{2}\sin \theta +\frac{\sqrt{3}}{2}\cos \theta \ge \frac{1}{2}$
or $\cos (\theta -\frac{\pi }{6})\ge \cos \left(\frac{\pi }{3}\right)$
or $-\frac{\pi }{3}\le \theta -\frac{\pi }{6}\le \frac{\pi }{3}$ $[\because -\pi <\theta \le \pi ]$
$\Rightarrow -\frac{\pi }{6}\le \theta \le \frac{\pi }{2}$