$\arccos(1/2)$ products

Note that\begin{align}\tan(x)=2\sin(x)&\iff\frac{\sin(x)}{\cos(x)}=2\sin(x)\\&\iff\sin(x)=0\vee\cos(x)=\frac12.\end{align}And$$\cos(x)=\frac12\iff x=\frac\pi3+2n\pi\vee x=-\frac\pi3+2n\pi,$$with $n\in\Bbb Z$.

In particular, although $\arccos\left(\frac12\right)\left(=\frac\pi3\right)$ is indeed a solution of the equation $\cos(x)=\frac12$, it is not the only one.

Let $ x\in[-\frac{\pi}{3},\frac{\pi}{3}] $.

$$\tan(x)=2\sin(x) \iff$$

$$\sin(x)(1-2\cos(x))=0\iff$$

$$\cos(x)=\frac 12 =\cos(\frac{\pi}{3})\iff $$

$$x=\pm \frac{\pi}{3}+2k\pi$$

but, if $ k\ne 0$, then $ x $ is out of the domain $ [-\frac{\pi}{3},\frac{\pi}{3}]$. So, $ k=0 $ and$$x=\pm \frac{\pi}{3}$$

The Cosine function is positive is Quadrants I and IV. Therefore, given $\cos(x)=\dfrac{1}{2}$, $x=\dfrac{\pi}{3}$ is the solution from Quadrant I and $x=-\dfrac{\pi}{3}$ (or $x=\dfrac{5\pi}{3}$) is the solution from Quadrant IV.

In high school, I remembered this using the acronym "All Students Take Calculus".