Solve the equation $2^{1-x} + 2^{\sqrt{2x-x^2}}=3 $ 
Solve the equation $$2^{1-x} + 2^{\sqrt{2x-x^2}}=3 \tag 1$$ on reals,
  using elementary knowledge (using trigonometry or logarithms is allowed, but without limits, differential calculus etc.)


We have to find solutions on $[0,2]$ interval.
Two solutions are easy to spot $x=0, x=1$
By rewriting (1) we get: $$2^{1-x} =3-2^{\sqrt{2x-x^2}} \tag 2$$
The left side is decreasing. The right side is decreasing on $[0,1]$ and increasing on $[1,2]$. It follows that there are no solutions in $(1, 2]$. 
The only interval I cannot cover is (0, 1).
Any help is appreciated.
 A: To begin with a seemingly random remark, note that 
$$4(1-s^2)\gt(2-s)^2\quad\text{for }0\lt s\le{1\over\sqrt2}\approx0.707$$
This is because the inequality simplifies to $s(4-5s)\gt0$, which holds for $0\lt s\lt{4\over5}=0.8$.
Now down to business.  As noted by Takahiro Waki in comments, the equation $2^{1-x}+2^{\sqrt{2x-x^2}}=3$ is equivalent to
$$2^{\sin\theta}+2^{\cos\theta}=3$$
with $-{\pi\over2}\le\theta\le{\pi\over2}$.  It's easy to see that $2^{\sin\theta}+2^{\cos\theta}\lt3$ for $-{\pi\over2}\le\theta\lt0$, and equality is achieved at $\theta=0$ and $\theta={\pi\over2}$ (corresponding to $x=1$ and $x=0$, respectively).  So it remains to show 
$$2^{\sin\theta}+2^{\cos\theta}\gt3\quad\text{for }0\lt\theta\lt{\pi\over2}$$
Actually, by the symmetry $\sin({\pi\over2}-\theta)=\cos\theta$ (and vice versa), we need only prove it in the interval $0\lt\theta\le{\pi\over4}$.  Now let's abbreviate this to
$$2^s+2^c\gt3$$
where $0\lt s\le{1\over\sqrt2}$ and $c=\sqrt{1-s^2}$.  To prove this, let's make clever use of AGM:
$$2^s+2^c=2^s+2^{c-1}+2^{c-1}\ge3\sqrt[3]{2^{s+2c-2}}$$
Thus we need only prove $s+2c-2\gt0$ for $0\lt s\le{1\over\sqrt2}$.  But this is straightforward: Since $c\ge0$, we have
$$\begin{align}
s+2c-2\gt0
&\iff2c\gt(2-s)\\
&\iff4c^2\gt(2-s)^2\\
&\iff4(1-s^2)\gt(2-s)^2
\end{align}$$
which takes us back to the obviously now non-random remark at the beginning.
A: The function $f(x)=2^{1-x} + 2^{\sqrt{2x-x^2}}$ has as domain $[0,2]$ and the derivative is $f'(x)= \log 2(\frac{2^{\sqrt{2x-x^2}}(x-1)}{\sqrt{2x-x^2}}-2^{1-x})$.
$f$ has a maximun at $x_m$, $f(x_m)\gt 3$ and $f(2)=\frac 32$, furthermore $f$ is decreasing in the interval $[x_m,2]$.
It follows since $f(0)=3$ there is an unique other value in the domain $[0,2]$ for which $f(x)=3$. This value is apparent and equal to $1$.
Thus the only soltions are $\color{red}0$ and $\color{red}1$.
