Solving $\cos^{-1}x+\cos^{-1}2x=-\pi$ gives the invalid answer $x=0$ To solve the inverse trigonometric equation
$$\cos^{-1}x+\cos^{-1}2x=-\pi,$$
I use the normal cosine addition
\begin{align}
\cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&=-\pi\\
2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}&=\cos(-\pi)\\
2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}&=-1\\
(2x^2+1)^2&=(1-x^2)(1-4x^2)\\
4x^4+1+4x^2&=1-4x^2-x^2+4x^4\\
9x^2&=0\\
x&=0.
\end{align}
Putting $x=0$ in the equation gives LHS $\ne$ RHS:
$$\cos^{-1}0+\cos^{-1}0=-\pi \\\pi=-\pi.$$
Since the equation has no solution, why does solving it give zero as a solution? Is my method wrong or is there something else which gives up one solution (i.e. $x=0$ on solving algebraically)?
 A: 
\begin{align}
\cos^{-1}x+\cos^{-1}2x=-\pi\tag{*}\\
\cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&=-\pi\tag1\\
2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}&=-1\tag2\\
(2x^2+1)^2&=(1-x^2)(1-4x^2)\tag3\\
4x^4+1+4x^2&=1-4x^2-x^2+4x^4\tag4\\
x&=0\tag5.
\end{align}

Note that $$\cos^{-1}x+\cos^{-1}2x=\begin{cases}2\pi-&\cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&\text{for }x\in[-0.5,0)\\&\cos^{-1}(2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2})&\text{for }x\in[0,0.5]\end{cases},$$ so step $(1)$ is actually invalid. Fortunately—replacing step $(1)$—the following argument/working is valid:
$$\begin{align}
&\cos^{-1}x+\cos^{-1}2x =-\pi\tag{*}\\
&\implies \cos(\cos^{-1}x+\cos^{-1}2x) =\cos(-\pi)\\
&\implies 2x^2 -\sqrt {1-x^2} \sqrt{1-4x^2}=-1\tag2\\
&\implies (2x^2+1)^2=(1-x^2)(1-4x^2)\tag3\\
&\implies 4x^4+1+4x^2=1-4x^2-x^2+4x^4\tag4\\
&\implies x=0\tag5.
\end{align}$$
Therefore, $$\text{equation } (*)\implies x=0.$$
Now, as pointed out by Crostul, since $\arccos$ is a nonnegative function, the given equation $(*)$ is trivially inconsistent (has no solution). Thus, $x{=}0$  is an extraneous solution (it was created in step $(2)\,).$

Since the equation has no solution, why does solving it give zero as a solution?

When a given equation $(A)$ implies equation $(B),$ $(A)$'s solution set (containing the actual solutions) is a subset of $(B)$'s solution set (containing the candidate solutions).
In particular, none of the candidate solutions being an actual solution is necessarily due to deductive explosion, which is necessarily due to equation $(A)$ being inconsistent.
