$\arccos(1/2)$ products
I'm finding points of intersection between two functions, $y = \tan(x)$ and $y = 2\sin(x)$, which are bound between $-\pi/3$ and $\pi/3$.
Solving for the intersections points, I end up performing $x = \arccos\left(\frac12\right)$
I know this produces: $x = \pi/3$, but should it rather produce $x = \pm\pi/3$ ?
This seems to make sense as visually the two functions intersect at both $-\pi/3$ and $\pi/3$, but I want to verify if this is correct.
I'm usually on Stack Overflow. Let me know if any notation here is wrong. Additional apologies if this question is very low level, I haven't worked with trig in a while.
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$\begingroup$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.
$\endgroup$ 2 $\begingroup$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$$x=\pm \frac{\pi}{3}$$
$\endgroup$ 2 $\begingroup$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".
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