If [2 sin x] + [cos x] = –3, then the range of the function $\mathrm{f}\left(\mathrm{x}\right)=\sin \mathrm{x}+\sqrt{3}\cos \mathrm{x}\text{ in }[0,2\mathrm{\pi }]$ (where [⋅] denotes the greatest integer function) is

We have, [2 sin x] + [cos x] = –3
only if [2 sin x] = –2 and [cos x] = –1
⇒ –2 ≤ 2 sin x < –1 and –1 ≤ cos x < 0
$\begin{array}{l}\Rightarrow -1\le \sin \mathrm{x}<-\frac{1}{2}\text{ and }-1\le \cos \mathrm{x}<0\\ \Rightarrow \frac{7\mathrm{\pi }}{6}<\mathrm{x}<\frac{11\mathrm{\pi }}{6}\text{ and }\frac{\mathrm{\pi }}{2}<\mathrm{x}<\frac{3\mathrm{\pi }}{2}\\ \Rightarrow \frac{7\mathrm{\pi }}{6}<\mathrm{x}<\frac{3\mathrm{\pi }}{2}\end{array}$
For the above values of x, $\sin \mathrm{x}+\sqrt{3}\cos \mathrm{x}=2\sin \left(\frac{\mathrm{\pi }}{3}+\mathrm{x}\right)$ lies between –2 and –1.
$\therefore$ Range of f (x) is (–2, –1).