Solving this trigonometric equation I want to solve this equation 
$$
\arccos(x)+\arcsin(x^2-x+1)=\pi/2
$$
I write this: for all $x\in [-1,1]$ 
$\arcsin(x^2-x+1)=\pi/2-\arccos(x)$ then $x^2-x+1=\sin(\pi/2-\arccos(x))=\cos(\arccos(x))=x$
$x^2-x+1=x\Rightarrow x^2-2x+1=0\Rightarrow (x-1)^2=0\Rightarrow x=1 $
Is it true ?
 A: What you wrote is almost correct.
I say almost because:


*

*$\arcsin$ map is only defined on $[-1,1]$. Hence you have to restrict to $x \in [0,1]$ in order to have $x^2-x+1 \in [0,1]$.

*You proceeded by implication. You should validate that $x=1$ is indeed a solution of the equation.


Apart from that your computations are correct.
A: By definition $\arccos(t)+\arcsin(t)=\dfrac\pi2$, implying
$$\arccos(x)=\frac\pi2-\arcsin(x^2-x+1)=\arccos(x^2-x+1)$$ 
and the arc cosine function, from $[-1,1]$ to $[0,\pi]$, is invertible so that this strictly equivalent to
$$x=x^2-x+1,x\in[-1,1]$$
which has the double root $x=1$.
A: Yes true indeed, in fact doubly true ( double root) at $ x=1$ 
Compare terms of two equations with identity
$$ \arccos(x)+\arcsin(x^2-x+1)=\pi/2 $$
$$ \arccos(x)+\arcsin(x)=\pi/2 $$
so we must have
$$ x^2-x+1 = x,\, x^2-2x+1 = 0,\,(x-1)^2=0,\,x= (1,1)$$
Solution can be verified by their graphs at  $(x=1)$ their average value is $y = \pi/4$ at $ x=1$ has a tangent point:
$$ \dfrac{\arccos(x)+\arcsin(x^2-x+1)}{2}=\pi/4 $$

It can be seen that $\arccos(x) $  has domain $(-1,1).$ 
However,  $\arcsin(x^2-x+1)$ and hence the sum $\arccos(x) + \arcsin(x^2-x+1)$ must have a restricted domain $(0,1)$.
When $x<0,\,\arcsin(x^2-x+1)$ is imaginary !
