How prove this equation has only one solution $\cos{(2x)}+\cos{x}\cdot\cos{(\sqrt{(\pi-3x)(\pi+x)}})=0$ 
Let $x\in (0,\dfrac{\pi}{3}]$.
  Show that this equation 
  $$\cos{(2x)}+\cos{x}\cdot\cos{(\sqrt{(\pi-3x)(\pi+x)}})=0$$
  has a unique solution $x=\dfrac{\pi}{3}$

I try to the constructor $$f(x)=\cos{(2x)}+\cos{x}\cdot\cos{(\sqrt{(\pi-3x)(\pi+x)}})=0\, , \quad\quad  f(\dfrac{\pi}{3})=0$$but I use found this function is not  a monotonic function
see wolframpha
Now the key How to prove this function $f(x)$ in $(0,\frac{\pi}{3})$ has no solution, since
$$f\left(\frac{\pi}{6}\right)=\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cos{\left(\dfrac{1}{2}\sqrt{\dfrac{7}{3}}\pi\right)}=-0.138\cdots<0$$
in other words, how to prove that
$$f(x)<0,\forall x\in(0,\dfrac{\pi}{3}) \, .$$

Not a solution, only a change of the question.
$\displaystyle a:=\sqrt{\frac{\pi-x}{2}+\sqrt{(\frac{\pi-x}{2})^2-x^2}}$ , $\enspace \displaystyle b:=\sqrt{\frac{\pi-x}{2}-\sqrt{(\frac{\pi-x}{2})^2-x^2}}$ 
=> $\enspace ab=x$ , $a^2+b^2=\pi-x$ , $a^2-b^2=\sqrt{(\pi-3x)(\pi+x)}$ 
The problem $f(x)=0$ with $\displaystyle 0<x<\frac{\pi}{3}$ changes to 
$\displaystyle \cos(2ab)=\cos(a^2+b^2)\cos(a^2-b^2)=\frac{1}{2}(\cos(2a^2)+\cos(2b^2)) \enspace$ or converted
$\cos(2a^2)+\cos(2b^2)-2\cos(2ab)=0$ .  
With $\enspace \alpha:=2a^2$ , $\beta:=2b^2\enspace $ and because of $\enspace 2ab=2\pi-\alpha-\beta\enspace $ one gets the condition 
$\cos(\alpha)+\cos(\beta)-2\cos(\alpha+\beta)=0 \enspace $ and $\enspace \alpha\beta=(2\pi-\alpha-\beta)^2\enspace$ with $\enspace\displaystyle 0<\beta<\frac{2\pi}{3}<\alpha<2\pi$ . E.g. :
$\displaystyle (\alpha;\beta)=(\pi;\arctan\frac{1}{3})$ is a solution but we get 
$3.65<(\pi-\beta)^2<3,7<\pi\beta<3.87\enspace$ with it.
Generally we have to show $\enspace \alpha\beta\neq (2\pi-\alpha-\beta)^2\enspace$ for all solutions of 
$\cos(\alpha)+\cos(\beta)-2\cos(\alpha+\beta)=0 \enspace$ in the given value range.
