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
$$f(x)<0,\forall x\in(0,\dfrac{\pi}{3}) \, .$$
 A: 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.
A: Let $f$ be defined on $(0,\dfrac{\pi}{3}]$ such that
$$f(x)=\cos(2x)+\cos(x)\cdot\cos\left(\sqrt{(\pi-3x)(\pi+x)}\right)$$
It is immediate that  the domain of the function could be $-\pi\le x\le\dfrac{\pi}{3}$ and that $f(-\pi)=f(-\frac{2\pi}{3})=f(0)=f(\frac{\pi}{3})=0$ (which is easily get from the radical) but the three values $x=-\pi,-\frac{2\pi}{3},0$ has been discarded by convention of the considered domain. It remains to verify that the only point of $(0,\dfrac{\pi}{3}]$ such that $f(x)=0$ is $x=\dfrac{\pi}{3}$.
Because of $\cos (2x)=2\cos^2(x)-1$ we have
$$f(x)=\cos(x)\left(2\cos(x)+\cos\left(\sqrt{(\pi-3x)(\pi+x)}\right)\right)-1$$
Consider the function defined on $0\lt x\lt\dfrac{\pi}{3}$
$$h(x)= 2\cos(x)+\cos\left(\sqrt{(\pi-3x)(\pi+x)}\right)$$ 
Taking derivative $$h’(x)=-2\sin x+\frac{(\pi+3x)\sin(\sqrt{(\pi-3x)(\pi+x)})}{\sqrt{{(\pi-3x)(\pi+x)}}}$$ It follows $$\begin{cases}h'(x)\lt0\space \text{for }0\lt x\lt 0.3711\Rightarrow h(x)\text{ is decreasing in the interval}\space (0,0.3711)\\h(0.3711)\approx0.9734\text{ is a minimun of h and }\cos(0.3711)\approx0.931929\\ h’(x)\gt 0\text{ for } x\gt0.3711;\space\space h(x_0)=1\text { for } x_0\approx0.538\text { and }\cos(0.538)\approx 0.858735 \end {cases}$$
Hence $h(x)\lt1$ for $0\lt x\lt0.538$ so $$\cos(x)h(x)\lt1\text{ on } 0\lt x\lt0.538$$ Consequently $$\color{red}{ f(x)=\cos(x)h(x)-1\lt0\text{ on } 0\lt x\lt0.538}\qquad(*)$$
It remains to prove the inequality for  $0.538\lt x\lt \dfrac{\pi}{3}$
$$\begin{cases}h'(x)\lt0\text{ when }\space 0.3711\lt x\lt\dfrac{\pi}{3}\Rightarrow h(x)\text{ increasing on }\space (0.538,\space \dfrac{\pi}{3})\\h(0.538)\approx 1\text{ and }h(\dfrac{\pi}{3})=2\end{cases}$$
Consider now the function $k(x)=\cos(x)h(x)$. Taking into account that in the interval $I=(0.538,\space \dfrac{\pi}{3})$ the function  $h(x)$ is increasing but $\cos(x)$ is decreasing from $\cos(0.538)\approx 0.858735$ till $\cos(\dfrac{\pi}{3})=\dfrac 12$ we can not conclude that k (x) is increasing in all the interval $I$.
Anyway we calculate the minimun of $k(x)$ on $I$; this corresponds to a unique root   of the equation $\cos(x)h'(x)=\sin(x)h(x)$ which is $x_0\approx 0.6439$ giving the minimun $k(x_0)\approx0.8481\gt0.$
It follows $$k(x)\ge k(x_0)\gt 0\text{ in the interval }[0.6439,\space\dfrac{\pi}{3})$$ hence $$0.8441\le k(x)\lt k\left(\dfrac{\pi}{3}\right)=\cos\left(\dfrac{\pi}{3}\right)h\left(\dfrac{\pi}{3}\right)=\dfrac 12\cdot 2=1$$
Consequently $$\color{red}{f(x)=k(x)-1\lt 0\text { on }[0.6439,\space\dfrac{\pi}{3})}\quad(**)$$
Thus, by $(*)$ and $(**)$, $\dfrac{\pi}{3}$ is the only root of $f(x)$ such that $0\lt x\le \dfrac{\pi}{3}$.  
