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First I solved it like this

$\arcsin x=\arccos x \iff \sin(\arcsin x)=\sin(\arccos x)$ implies for $x\in[-1,1]$:

$x=\pm\sqrt{1-x^2} \implies x^2=1-x^2\iff x=\pm \frac 1{\sqrt 2}$ but $-\frac 1{\sqrt 2}\in[-1,1]$ is not solution, what is the mistake?

Did you detect the error by checking? as in irrational equations ?

However

$\arcsin x=\arccos x \iff \arcsin x=\frac{\pi}2-\arcsin x\iff 2\arcsin x=\frac{\pi}2 \iff \arcsin x=\frac{\pi}4$ implies for $x\in[-1,1]$:

$x=\sin \frac{\pi}4=\frac1{\sqrt 2}$.

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    $\begingroup$ There is no mistake (except that $\sin(\arccos x)=\sqrt{1-x^2}):$ squaring an equation will sometimes create extraneous solutions (and you've carefully used and in all the right places). $\endgroup$
    – ryang
    Aug 4 at 17:44
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    $\begingroup$ $\sin(\arccos(x))=\sqrt{1-x^2}\ge0$, so $x\ge0$. Arccos(x) is an angle between $0$ and $\pi$ so the sine of such an angle cannot be negative. $\endgroup$ Aug 4 at 17:46
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    $\begingroup$ @ryang thanks, I hadn't noticed $\endgroup$
    – Pierre
    Aug 4 at 17:47

1 Answer 1

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Actually, for each $x\in[-1,1]$, $\sin(\arccos x)=\sqrt{1-x^2}$. Obviously, if $x<0$, the equation $x=\sqrt{1-x^2}$ has no solutions. And, if $x\geqslant0$,\begin{align}x=\sqrt{1-x^2}&\iff x^2=1-x^2\\&\iff x^2=\frac12\\&\iff x=\frac1{\sqrt2}\end{align}(since $x\geqslant0$).

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