<s>[INST] Solve the following trigonometric inequalities I'm solving the following inequality:
$$\cos(2x+\frac \pi 3) \ge - \frac 1 2 \text{ with } -\frac \pi 2<x<\frac \pi 2$$
I found the following range of solutions:
$$\frac 4 3 \pi \leq2x+\frac \pi 3 \leq \frac 2 3 \pi$$
That I simplified with
$$\frac \pi 3 \leq 2x\leq\pi \rightarrow \frac \pi 6 \leq 2x<\frac \pi 2$$
Though, my book suggests the following solutions:
$$-\frac \pi 2< x \leq \frac \pi 6$$
Any hints on why my solution is wrong?
 [/INST] Let $\theta = 2x + \dfrac{\pi}{3}$.
Observe that the requirement that $x$ satisfies the inequalities 
$$-\frac{\pi}{2} < x < \frac{\pi}{2}$$
implies that 
$$\theta = 2x + \frac{\pi}{3} > 2\left(-\frac{\pi}{2}\right) + \frac{\pi}{3} = -\pi + \frac{\pi}{3} = -\frac{2\pi}{3} \tag{1}$$
and that
$$\theta = 2x + \frac{\pi}{3} < 2\left(\frac{\pi}{2}\right) + \frac{\pi}{3} = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \tag{2}$$
Combining inequalities 1 and 2 yields 
$$-\frac{2\pi}{3} < \theta < \frac{4\pi}{3}$$
Hence, the inequality 
$$\cos\left(2x + \frac{\pi}{3}\right) \geq -\frac{1}{2}, -\frac{\pi}{2} < x < \frac{\pi}{2} \iff \cos\theta \geq -\frac{1}{2}, -\frac{2\pi}{3} < \theta < \frac{4\pi}{3}$$
In the interval $(-2\pi/3, 4\pi/3)$, 
$$\cos\theta \geq -\frac{1}{2} \implies -\frac{2\pi}{3} < \theta \leq \frac{2\pi}{3}$$
Substituting $2x + \dfrac{\pi}{3}$ for $\theta$ yields
\begin{align*}
-\frac{2\pi}{3} & < 2x + \frac{\pi}{3} \leq \frac{2\pi}{3}\\
-\pi & < 2x \leq \frac{\pi}{3}\\
-\frac{\pi}{2} & < x \leq \frac{\pi}{6}
\end{align*}
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