Question

$\mathrm{The}\mathrm{range}\mathrm{of}\theta \mathrm{for}\mathrm{which}\mathrm{the}\mathrm{inequality}\sin \theta +\sqrt{3}\cos \theta \ge 1\mathrm{is}\mathrm{valid}\mathrm{if}\theta \in \left(-\pi ,\pi \right]\mathrm{is}$

• A. $\theta \in \left(-\frac{\pi }{3},\frac{\pi }{2}\right]$
• B. $\theta \in \left(-\frac{\pi }{6},\frac{\pi }{2}\right]$
• C. $\theta \in \left(\frac{\pi }{3},\frac{\pi }{2}\right]$
• D. $\theta \in \left(\frac{\pi }{6},\frac{\pi }{2}\right]$

Answer 1

The correct option is B $\theta \in \left(-\frac{\pi }{6},\frac{\pi }{2}\right]$

Here, the given inequation
$\sin \theta +\sqrt{3}\cos \theta \ge 1$ can solved as:
$\sin \theta +\sqrt{3}\cos \theta \ge 1\mathrm{Dividing}\mathrm{by}2\mathrm{in}\mathrm{both}L.H.S\mathrm{and}R.H.S\mathrm{we}\mathrm{get},\frac{1}{2}\sin \theta +\frac{\sqrt{3}}{2}\cos \theta \ge \frac{1}{2}\Rightarrow \sin \left(\frac{\pi }{6}\right)\sin \theta +\cos \left(\frac{\pi }{6}\right)\cos \theta \ge \frac{1}{2}\Rightarrow \cos \left(\theta -\frac{\pi }{6}\right)\ge \frac{1}{2}=\cos \left(\frac{\pi }{3}\right)\Rightarrow -\frac{\pi }{3}\le \theta -\frac{\pi }{6}\le \frac{\pi }{3}\Rightarrow -\frac{\pi }{6}\le \theta \le \frac{\pi }{2}.$