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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.

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  • $\begingroup$ If you square an equation, you can make sure you don't have extraneous solutions simply by plugging in all the values you found and discarding the ones that don't solve the equation. $\endgroup$ – Cheerful Parsnip Jan 18 '18 at 22:26
  • $\begingroup$ The addition equation $x =2$ didn't come from squaring $2x-x^2$. I think it came from evaluating $\cos(sin^{-1}(1-x)$ as $\sqrt {2x - x^2}$. I think. $\endgroup$ – fleablood Jan 18 '18 at 22:32
  • $\begingroup$ @GrumpyParsnip that is the most obvious thing to do. but, question was how to avoid extra solutions without doing it. $\endgroup$ – ss1729 Jan 18 '18 at 22:35
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    $\begingroup$ @ss1729 I realize it's obvious, but it's a perfectly valid method, and I wanted to make sure people realized that. The main issue here is that when you square, you are proving a forward implication, whereas you really need an iff. But the fact that you have a forward implication is enough to show that the set of correct solutions is a subset of the ones you find. $\endgroup$ – Cheerful Parsnip Jan 18 '18 at 22:41
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    $\begingroup$ Taking cosine can also introduce invalid solutions. Whenever you apply a non-injective function to both sides of an equation, this can happen. And it's quite common, and perfectly valid, to only check in the end by plugging solutions into the original equation. $\endgroup$ – Jean-Claude Arbaut Jan 19 '18 at 8:18
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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 $x=0$.

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  • $\begingroup$ thnx. i didnt think that way. though i am always afraid of squaring while solving such equations. $\endgroup$ – ss1729 Jan 18 '18 at 22:18
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    $\begingroup$ Just remember that $x^2=y^2\Leftrightarrow x=y$ for $xy\geq0$. $\endgroup$ – Michael Rozenberg Jan 18 '18 at 22:22
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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.

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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?

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