# Solve $\arcsin x=\arccos x, x\in[-1,1]$

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 $$x\in[-1,1]$$:

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

• 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). Aug 4 at 17:44
• $\sin(\arccos(x))=\sqrt{1-x^2}\ge0$, so $x\ge0$. Arccos(x) is an angle between $0$ and $\pi$ so the sine of such an angle cannot be negative. Aug 4 at 17:46
• @ryang thanks, I hadn't noticed Aug 4 at 17:47

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 $$x\geqslant0$$).