How to solve: $\arccos(2x)-\arccos(x)=\pi/3$ Solve for $x$ the equation $\arccos(2x)-\arccos(x)=\pi/3$.
My attempt using $\cos(a-b)$. But it gives me a sqrt-expression that I don't know how to handle.
WolframAlpha
 A: This reduces to an easy quadratic.  First, take the cosine of both sides. Using the cosine addition formula, I get
$$2 x \cdot x + \sqrt{1-4 x^2} \sqrt{1-x^2} = \frac12$$
Manipulate and square both sides to get
$$\left ( 2 x^2-\frac12\right)^2 = (1-4 x^2)(1-x^2) = 1-5 x^2+4 x^4$$
or
$$3 x^2=\frac{3}{4} \implies x = \pm \frac{1}{2}$$
Now plugging in both answers, one sees that only the $-1/2$ result makes sense for the principal branch (or any branch) of the arccos.  Thus, $x=-1/2$.
A: Hint: $\arccos\left(2x\right)=\arccos\left(x\right)+\pi/3$. Now take the
cosinus on both sides and see what happens.
A: Avoid squaring which often incurs the burden of extraneous roots
Let  $\displaystyle \arccos x=\theta\ \ \ \ (1)\implies x=\cos\theta$
We have $\displaystyle \arccos(2\cos\theta)-\theta=\frac\pi3\implies \arccos(2\cos\theta)=\theta+\frac\pi3\ \ \ \ (2) $
As the  principal value of $\arccos$ lies $\in[0,\pi],$ 
from $(1), 0\le\theta\le\pi$
form $(2), 0\le\theta+\frac\pi3\le\pi\implies -\frac\pi3\le\theta\le\frac{2\pi}3$
So, we need $\displaystyle 0\le \theta\le \frac{2\pi}3\  \ \  \ (3)$
$\displaystyle (2)\implies 2\cos\theta=\cos\left(\frac\pi3+\theta\right)=\cos\frac\pi3\cos\theta-\sin\frac\pi3\sin\theta$
$\displaystyle \implies\frac{\sqrt3}2\sin\theta=\left(\frac12-2\right)\cos\theta$
$\displaystyle \implies\tan\theta=-\sqrt3=\tan\left(-\frac\pi3\right)$
$\displaystyle \implies \theta=n\pi-\frac\pi3$ where $n$ is any integer
From $\displaystyle(3), 0\le n\pi-\frac\pi3\le \frac{2\pi}3\iff 0\le3n-1\le2\implies n=1$
Can you take it from here?
