The equation with arccos and arcsin For what value of parameter $a$ the equation
$$ \text{arccos}(\sin(x)+a)=\text{arcsin}(\cos(x)+a) $$  
has exactly one solution in the interval for $x$ $(0; 2\pi)$?
 A: Suppose that the equation has only one solution, say $x_0$. Let
$$\alpha:=a+\,\sin x_0\\ \beta:=a+\cos x_0$$
then after taking sine and cosine from the both sides of your equation:
$$\sin(\arccos{\alpha})=\beta\\ \cos(\arcsin{\beta})=\alpha$$
So we can write:
$$\begin{align}\alpha^2+\beta^2&=\sin^2(\arccos{\alpha})+\cos^2(\arcsin{\beta})
\\ &=1-\alpha^2+1-\beta^2\\ \implies \alpha^2+\beta^2&=1
\end{align}$$
On the other hand:
$$\begin{align}\alpha^2+\beta^2 &=(a+\sin x_0)^2+(a+\cos x_0)^2\\
&=1+2a^2+2a(\sin x_0+\cos x_0)
\end{align}$$
Therefore
$$1+2a^2+2a(\sin x_0+\cos x_0)=1\\
\implies a=-\sin x_0-\cos x_0\quad\text{or}\quad a=0$$
Now you can see that for $a=0$ every $x$ in $(0,\frac{\pi}2)$ satisfies the equation. So it is safe to say that if the equation has only one solution in $x_0$ then we must have
$$a=-\sin x_0-\cos x_0=\sqrt{2}\cos(x_0+3\pi/4)$$
Of course this is a necessary and not sufficient condition. Taking this into account, I plotted the following figure:

As you see, if $x_0\in[0,\,2\pi]$ satisfies your equation then we are sure that $\pi\le x_0\le \frac{3\pi}2$.
Since $a=\sqrt{2}\cos(x_0+3\pi/4)$ we can say that $a$ is a function of $x_0$ whose domain is $[\pi,\frac{3\pi}{2}]$. Also it is easy to verify that the range of this function is $[1,\,\sqrt{2}]$. So in conclusion:


*

*If $0<a<1$ or $a>\sqrt{2}$ then your equation has no solution in $[0,\;2\pi]$.

*If $1\le a\le\sqrt{2}$ then the solution of your equation is equivalent to the solution of
$$\cos(x+3\pi/4)=\frac a{\sqrt 2}$$
which has two solutions for $a<\sqrt 2$ and only one for $a=\sqrt 2$.

