<s>[INST] Trigonometric equation 3 $$2 \arcsin x = \arccos 2x$$
I tried applying $\sin$ to the equation, so it results
$$\sin (2 \arcsin x) = \sin(\arccos 2x)$$
$$2 \sin(\arcsin x) \cdot \cos(\arcsin x) = \sin (\arccos 2x)$$
$$2x \cdot \sqrt{1 - x^2} = \sqrt{1- 4x^2}$$
But this doesn't result in the correct answer $\frac{\sqrt{3} -1}{2}$
What am I missing ?
 [/INST] NOTE: I would suggest doing $\cos(2\arcsin x)=\cos(\arccos 2x)$ instead which seems easier since $\cos(2\arcsin x)=1-2\sin^2(\arcsin x)=1-2x^2$
About your method you're not missing anything
$$(2x)^2(1-x^2)=1-4x^2\\4x^2-4x^4=1-4x^2\\4x^4-8x^2+1=0\\x^2=t\\4t^2-8t+1=0\\t_{1,2}=\frac{8\pm\sqrt{64-16}}{8}=\frac{8\pm4\sqrt{3}}{8}=\frac{2\pm\sqrt{3}}{2}$$
Now we can eliminate the $\frac{2+\sqrt3}{2}$ since then $x>1$ so the square root 
of $1-x^2$ is not defined.
$$t=\frac{2-\sqrt3}{2}=\frac{4-2\sqrt3}{4}=\frac{(\sqrt 3-1)^2}{4}\\x^2=\frac{(\sqrt 3-1)^2}{4}\\x=\pm\frac{\sqrt 3-1}{2}$$
We can rule out the negative solution since then $2x\sqrt{1-x^2}$ is negative while $=\sqrt{1-4x^2}$ is positive.
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