Mathematics - Trigonometric Equation Question with Solution | TestHub

The inequality $(2\sin x-\sqrt{3})(2\cos x-1)\ge 0$ holds if (1) $\sin x\ge \frac{\sqrt{3}}{2}$ and $\cos x\ge \frac{1}{2}$, or (2) $\sin x\le \frac{\sqrt{3}}{2}$ and $\cos x\le \frac{1}{2}$. In $[0,2\pi ]$: (1) $\sin x\ge \frac{\sqrt{3}}{2}$ implies $x\in [\pi /3,2\pi /3]$, and $\cos x\ge \frac{1}{2}$ implies $x\in [0,\pi /3]\cup [5\pi /3,2\pi ]$. Their intersection is $x=\pi /3$. (2) $\sin x\le \frac{\sqrt{3}}{2}$ implies $x\in [0,\pi /3]\cup [2\pi /3,2\pi ]$, and $\cos x\le \frac{1}{2}$ implies $x\in [\pi /3,5\pi /3]$. Their intersection is $x\in \{\pi /3\}\cup [2\pi /3,5\pi /3]$. Combining both cases, the solution set is $x\in \{\pi /3\}\cup [2\pi /3,5\pi /3]$. For integer values of $x$, $\pi /3\approx 1.047$ is not an integer. The interval $[2\pi /3,5\pi /3]$ is approximately $[2.094,5.236]$. The integers in this range are $3,4,5$. Thus, there are 3 integer values of $x$.