How to go about solving this inequality question? $\cos(3x-\pi/3) \leq (1/2).$ 
Here is what I have done so far...
Let $3x-\pi/3 = X$. So I need to solve $\cos(X) \leq 1/2$. Which is all $X$ from $\pi/3$ to $5\pi/3$, so--
$\pi/3 \leq X \leq 5\pi/3 \quad\longrightarrow\quad \pi/3 \leq 3x-\pi/3 \leq 5\pi/3.$
So should I now solve for $x$? or...

It's more correct to say that 
$$ \frac{\pi}{3} + 2n\pi \le 3x - \frac{\pi}{3} \le \frac{5\pi}{3} + 2n\pi, \quad n \in \mathbb{Z} $$
Solving for $x$:
$$ \frac{2\pi}{3} + 2n\pi \le 3x \le 2\pi + 2n\pi $$
$$ \frac{2}{9} + \frac{2n\pi}{3} \le x \le \frac{2\pi}{3} + \frac{2n\pi}{3} $$
or 
$$\frac{2 + 6n}{9}\,\pi \le x \le \frac{2 + 2n}{3}\,\pi$$
What this means is $x$ lies in not one interval but a family of intervals, equally spaced by $\frac{2\pi}{3}$. For example $\frac{2\pi}{9} \le x \le \frac{2\pi}{3}$ is a solution but $\frac{8\pi}{9} \le x \le \frac{4\pi}{3}$ and $\frac{14\pi}{9} \le x \le 2\pi$ are also solutions.
