Question

The value of $x$ for $\sin x+\sqrt{3}\cos x\ge 1$ in the interval $-\pi <x\le \pi$ is

• A. $-\frac{\pi }{2}\le x\le \frac{\pi }{6}$
• B. $-\frac{\pi }{6}\le x\le \frac{\pi }{2}$
• C. $0\le x\le \frac{\pi }{4}$
• D. $-\frac{\pi }{4}\le x\le \frac{\pi }{4}$

Answer 1

The correct option is B $-\frac{\pi }{6}\le x\le \frac{\pi }{2}$

The given inequation is
$\sin x+\sqrt{3}\cos x\ge 1$
$\Rightarrow \frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x\ge \frac{1}{2}\Rightarrow \cos (x-\frac{\pi }{6})\ge \frac{1}{2}$
Now, draw the graph.

In the above graph, intersection points are $-\frac{\pi }{6}$ and $\frac{\pi }{2}$
So, from the graph
$-\frac{\pi }{6}\le x\le \frac{\pi }{2}$