When does $\arcsin^2x+\arccos^2x=1$ for $x$ real or comlex? let $\arcsin$ is the compositional inverse of $\sin$ and $\arccos$ the compositional inverse of $\cos$ , my question here is it possible to find $x$ for which : $\arcsin^2x+ \arccos^2x=1$.
Note: $x$ is a real number or complex 

You want $u^2+v^2=1$ and $\sin u=\cos v$. This means:
Letting $z=e^{iv}$ you want $z+\frac{1}{z}=2\sin u$ or $z=\frac{2\sin u\pm \sqrt{4\sin^2 u-4}}{2}=\sin u \pm i\cos u$.
If $u=\frac{\pi}{2}-w$ then this is $z=e^{\pm iw}$ or $v=2\pi k \pm\left( \frac{\pi}{2}-u\right)$. 
Case 1: If $u+v=2\pi k +\frac{\pi}{2}=S_+$  then
$$uv = \frac{(u+v)^2-(u^2+v^2)}{2}=\frac{S_+^2-1}{2}$$
Giving that $u,v$ are roots of $x^2-S_+x+\frac{S_+^2-1}{2}=0$, or:
$$u,v=\frac{S_+\pm \sqrt{2-S_+^2}}{2}$$
Since $S_+$ is real an absolute value greater than $\sqrt{2}$, the real parts of $u,v$ must be $\frac{S_+}{2}.$
To get $u,v$ in the range of $\arcsin,\arccos$ you need the real part of $u$ to be in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ and you need the real part of $v$ to be in $[0,\pi].$ This means $S_+\in[0,\pi]$ or $k=0$. So we get two solutions:
$$x=\cos v=\cos\left(\frac{\pi \pm i\sqrt{\pi^2-8}}{4}\right)$$
Case 2: $v-u = 2\pi k - \frac{\pi}{2}=S_-$.
$$uv = \frac{u^2+v^2-(v-u)^2}{2}=\frac{1-S_-^2}{2}$$
So $v,-u$ are roots of $x^2-S_-x-\frac{1-S_-^2}{2}=0.$
And thus $$v,-u=\frac{S_-\pm \sqrt{2-S_-^2}}{2}$$
Now the real part of $u$ is $\frac{-S_-}{2}$ and the real part of $v$ is $\frac{S_-}{2}$, so you need $S_-\in[-\pi,\pi]$ and $S_-\in[0,2\pi]$ or $S_-\in[0,\pi]$. This is not possible with integer $k$.

If you take $\arccos$ and $\arcsin$ as multivalued on $\mathbb C$ then you get a more complicated result - basically, any $k$ gives two pairs $u,v$ for each case, (four pairs total) and then you get $x=\sin u=\cos v$. 
