Solve $\sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x$ and avoid extra solutions while squaring

Question

Solve the equation,

$$ \sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x $$

My Attempt: $$ \cos\Big[ \sin^{-1}x+\sin^{-1}(1-x) \Big]=x\\ \cos\big(\sin^{-1}x\big)\cos\big(\sin^{-1}(1-x)\big)-\sin\big(\sin^{-1}x\big)\sin\big(\sin^{-1}(1-x)\big)=x\\ \sqrt{1-x^2}.\sqrt{2x-x^2}-x.(1-x)=x\\ \sqrt{2x-x^2-2x^3+x^4}=2x-x^2\\ \sqrt{x^4-2x^3-x^2+2x}=\sqrt{4x^2-4x^3+x^4}\\ x(2x^2-5x+2)=0\\ \implies x=0\quad or \quad x=2\quad or \quad x=\frac{1}{2} $$ Actual solutions exclude $x=2$.ie, solutions are $x=0$ or $x=\frac{1}{2}$. I think additional solutions are added because of the squaring of the term $2x-x^2$ in the steps.

So, how do you solve it avoiding the extra solutions in similar problems ?

Note: I dont want to substitute the solutions to find the wrong ones.

Answer 1

The domain gives $$-1\leq x\leq1$$ and $$-1\leq1-x\leq1,$$ which gives $$0\leq x\leq1,$$ which says that the answer is $$\left\{\frac{1}{2},0\right\}.$$ I think it's better after your third step to write $$\sqrt{2x-x^2}=\sqrt{1-x^2}$$ or $x=0$.

Answer 2

Here's a way to avoid the extraneous solution. Note that $\arcsin u+\arccos u={\pi\over2}$ for all $u\in[-1,1]$. Thus we can rewrite $\arcsin x+\arcsin(1-x)=\arccos x$ as

$$\arcsin x+{\pi\over2}-\arccos(1-x)={\pi\over2}-\arcsin x$$

which simplifies to

$$2\arcsin x=\arccos(1-x)$$

Applying $\cos$ to each side and using $\cos(2\theta)=1-2\sin^2\theta$, we get $1-2x^2=1-x$, or

$$2x^2-x=0$$

which has $x=0$ and $x={1\over2}$ as its only solutions.

Answer 3

Like Barry Cipra,

$$2\arcsin x=\arccos(1-x)$$

Now $0\le\arccos(1-x)\le\pi$ and $-\pi\le2\arcsin x\le\pi$

$\implies\arcsin x\ge0\iff x\ge0$

Now for $\arcsin x\ge0,2\arcsin x=\begin{cases}\arccos(1-2x^2)&\mbox{if }x\ge0\\ -\arccos(1-2x^2)& \mbox{if }x<0\end{cases}$

$x\ge0\implies 1-x=1-2x^2$

Can you take it from here?