Two trig inequalities that must hold at once

Solve each inequality on $-\pi \le \theta \le \pi$, then keep only the angles in both.

Cosine: $2\cos \theta >-1$ means $\cos \theta >-\frac{1}{2}$. Cosine equals $-\frac{1}{2}$ at $\theta =\pm \frac{2\pi }{3}$ and is larger between them, so $$-\frac{2\pi }{3}<\theta <\frac{2\pi }{3}$$

Sine: $2\sin \theta >\sqrt{3}$ means $\sin \theta >\frac{\sqrt{3}}{2}$. Sine equals $\frac{\sqrt{3}}{2}$ at $\theta =\frac{\pi }{3}$ and $\theta =\frac{2\pi }{3}$ and is larger between them. Sine is negative for every angle in $(-\pi ,0)$, so nothing there qualifies: $$\frac{\pi }{3}<\theta <\frac{2\pi }{3}$$

Both together: the sine interval already sits inside the cosine interval, so the overlap is $$\frac{\pi }{3}<\theta <\frac{2\pi }{3}$$ That is choice D.

Each choice

• A. Cosine equals $-\frac{\sqrt{3}}{2}$ at $\pm \frac{5\pi }{6}$, so this solves $\cos \theta >-\frac{\sqrt{3}}{2}$: the special values for sine and cosine got swapped, and the sine condition was then dropped. It includes $\theta =0$, where $2\sin 0=0$ is not greater than $\sqrt{3}$.
• B. These angles solve $\sin \theta >\frac{1}{2}$, which is what you get by mixing up the values $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$. At $\theta =\frac{\pi }{4}$, for instance, $2\sin \theta =\sqrt{2}$, which is less than $\sqrt{3}$.
• C. This is the solution of the cosine inequality alone. It needs to be intersected with the sine condition, which removes everything below $\frac{\pi }{3}$.
• D (correct). The sine condition gives $\frac{\pi }{3}<\theta <\frac{2\pi }{3}$, and every angle there also has $\cos \theta >-\frac{1}{2}$.