Pre-calculus (solve the equation for exact solutions) Arccos x + arccos 2x = arccos 1/2 I have spent two hours working on this question and can’t find the answer. I believe I need to use the multiplication of cos, but don’t know how.
 A: Hint: Taking the cosine of both sides gives
$$
2x^2-\sqrt{1-x^2}\sqrt{1-4x^2}=\frac12
$$
This has two solutions, of which only one works in the original equation.
A: Taking $cos$ of both sides (both sides are some angles), we have $$cos(arccos(x)+arccos(2x))=cos(arccos(1/2))$$ and using identity for $cos$ of sum we get
$$cos(arccos(x))cos(arccos(2x))-sin(arccos(x))sin(arccos(2x))=1/2$$
or (using the fact that $sin \: x=\sqrt{1-cos^2(x)})$,
$$ 2x^2-\sqrt{1-x^2}\sqrt{1-4x^2}=1/2$$
Rearranging and squaring both parts we get $3x^2=3/4$, $x= 1/2$
A: $\arccos(2x)$ is defined if $-1\le2x\le1\iff-\dfrac12\le x\le\dfrac12\  \ \  \ (1)$
Again, if $x\le0,\arccos(2x),\arccos(x)\ge\dfrac\pi2$
But $\arccos(2x)+\arccos(x)=\arccos\dfrac12=\dfrac\pi3\implies x>0\  \ \  \ (2)$ and $\arccos x=+\arcsin\sqrt{1-x^2}$
Now $\arccos(2x)=\dfrac\pi3-\arccos(x)$
Now taking cosine in both sides,
$$2x=\cos\left(\dfrac\pi3-\arccos(x)\right)=\dfrac x2+\dfrac{\sqrt{3(1-x^2)}}2$$
$$\iff\dfrac{3x}2=\dfrac{\sqrt{3(1-x^2)}}2(\text{which is also }\ge0)$$
Square  both sides and check if $x$ satisfies $(1),(2)$ 
