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?
 A: 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*}
A: A good first step is to solve the inequality
$$
\cos(2x + \frac\pi3) \geq -\frac12
$$
without the additional restriction $-\frac\pi2 < x < \frac\pi2$.
It seems you attempted to do this, so that was good.
Let's write $u = 2x + \frac\pi3$ for a moment, because I think having
too many symbols in an inequality sometimes gets in the way of understanding.
We have to solve $\cos u \geq -\frac12$.
As you deduced, there are two points on the unit circle that have
$x$-coordinate $-\frac12$, and they occur at angles
$u = \frac23 \pi$ and $u = \frac 43 \pi$.
You also seem to have deduced that you want angles that are swept
out in the counterclockwise direction from the point
at $u = \frac 43 \pi$ to the point at $u = \frac23 \pi$.
Where you first went wrong was to use that conclusion to
construct the inequality
$$ \frac 43 \pi \leq u \leq \frac23 \pi.$$
Now that we have replaced the clutter of $2x + \frac\pi3$
with the single symbol $u$ in the middle of this chain of
inequalities, it may be more obvious that there is no value of $u$
that can possibly satisfy these inequalities;
if you could find one, you would have proved that
$\frac 43 \pi \leq \frac23 \pi$, which is not true.
In order to express a range of angles, you need the upper bound
of the range to be greater than the lower bound.
The way to accomplish this is to remember that each point on the
unit circle is identified by an infinite number of angle values;
for example, the point at angle $\frac43 \pi$ is also at angle
$2\pi + \frac43 \pi$, at angle $4\pi + \frac43 \pi$,
and at angle $-2\pi + \frac43 \pi$, among others.
Now, we could set the upper bound of the range to an angle equivalent to $\frac23 \pi$ and greater than $\frac43 \pi$;
or we could set the lower bound to an angle equivalent to $\frac43 \pi$ and less than $\frac23 \pi$.
I would choose the second option because I prefer angles closer to zero.
So we can try $-2\pi + \frac43 \pi = -\frac23 \pi$ as a lower bound,
that is, let the angle $u$ satisfy
$$ -\frac 23 \pi \leq u \leq \frac23 \pi.$$
Check this on a unit circle; it works.
The angle $-4\pi + \frac43 \pi = -\frac83 \pi$, although it is
equivalent to $\frac43 \pi$, would not be an acceptable choice, because from $-\frac83 \pi$ to $\frac23 \pi$ you travel all the way around the circle
and then some more, including angles $u$ that do not satisfy 
$\cos u \geq -\frac12$.
Now you can rewrite $-\frac 23 \pi \leq u \leq \frac23 \pi$
by replacing $u$ with its definition in terms of $x$:
$$ -\frac 23 \pi \leq 2x + \frac\pi3 \leq \frac23 \pi.$$
This is a correct version of your second displayed formula;
you should be able to get a correct solution from that.
