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 $$f(x)<0,\forall x\in(0,\dfrac{\pi}{3}) \, .$$

$$ \begin{align} & \text{Let}\quad f(x)=\cos{(2x)}+\cos{(x)}\cos{\left(\sqrt{3(\pi/3-x)(\pi+x)}\right)} \\ & \qquad (\pi/3-x)(\pi+x) \ge 0 \qquad \left\{\space x \in \left[ -\pi \,, +\pi/3 \right] \right\} \\ & \text{Substitute}\quad x=\frac{2\pi}{3}t-\frac{\pi}{3} \qquad \left\{\space \color{red}{t \space \in \left[ -1 \,, +1 \right]} \right\} \\ & \qquad \left( \pi/3-x \right) \left( \pi+x \right) = \left( \frac{2\pi}{3}-\frac{2\pi}{3}t \right) \left( \frac{2\pi}{3}+\frac{2\pi}{3}t \right) = \left( 2\pi/3 \right)^2 \left( 1-t^2 \right) \Rightarrow \\ & \qquad f(t)=\cos{\left[ \frac{2\pi}{3} \left( 1-2t \right) \right]}+\cos{\left[ \frac{\pi}{3} \left( 1-2t \right) \right]}\cos{\left[ \frac{2\pi}{\sqrt{3}} \sqrt{1-t^2} \right]} \\ & \text{Substitute}\quad \color{red}{t=\sin{\theta}} \quad\Rightarrow \sqrt{1-t^2}=\cos{\theta} \space\qquad \left\{\space \theta \in \left[ -\pi/2 \,, +\pi/2 \right] \right\} \\ & \qquad f(\theta)=\cos{\left[ \frac{2\pi}{3} \left( 1-2\sin{\theta} \right) \right]}+\cos{\left[ \frac{\pi}{3} \left( 1-2\sin{\theta} \right) \right]}\cos{\left[ \frac{2\pi}{\sqrt{3}} \cos{\theta} \right]} \\ & \text{Where}\space f(\theta) \space\text{is valid}\space\space \forall \space \theta \in \Bbb R \quad\&\quad f(\theta)=f(\theta+2\pi) \quad \text{is periodic} \\ & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \\ \end{align} $$
$$ \begin{align} & \text{Product-to-Sum:}\quad \cos{\left[ \frac{\pi}{3}-\frac{2\pi}{3}\sin{\theta} \right]}\cos{\left[ \frac{2\pi}{\sqrt{3}} \cos{\theta} \right]} = \\ & \small \qquad = \frac{1}{2} \cos{\left[ \frac{\pi}{3}-\frac{2\pi}{3} \left( \sin{\theta}+\sqrt{3}\cos{\theta} \right) \right]} + \frac{1}{2} \cos{\left[ \frac{\pi}{3}-\frac{2\pi}{3} \left( \sin{\theta}-\sqrt{3}\cos{\theta} \right) \right]} \\ & \small \qquad = \frac{1}{2} \cos{\left[ \frac{\pi}{3}-\frac{4\pi}{3}\cos{\left(\frac{\pi}{6}-\theta \right)} \right]} + \frac{1}{2} \cos{\left[ \frac{\pi}{3}+\frac{4\pi}{3}\cos{\left(\frac{\pi}{6}+\theta \right)} \right]} \normalsize \Rightarrow \\ & f(\theta) = \cos{\left[ \frac{2\pi}{3}-\frac{4\pi}{3}\sin{\theta} \right]} + \small \frac{1}{2} \cos{\left[ \frac{\pi}{3}-\frac{4\pi}{3}\cos{\left(\frac{\pi}{6}-\theta \right)} \right]} + \frac{1}{2} \cos{\left[ \frac{\pi}{3}+\frac{4\pi}{3}\cos{\left(\frac{\pi}{6}+\theta \right)} \right]} \\ & \text{Add-and-Sub:}\quad \small +\frac{1}{2} \cos{\left[ \frac{\pi}{3}+\frac{4\pi}{3}\cos{\left(\frac{\pi}{6}+\theta \right)} \right]}-\frac{1}{2} \cos{\left[ \frac{\pi}{3}+\frac{4\pi}{3}\cos{\left(\frac{\pi}{6}+\theta \right)} \right]} \normalsize \Rightarrow \\ & f(\theta) = \small \left( \cos{\left[ \frac{2\pi}{3}-\frac{4\pi}{3}\sin{\theta} \right]} + \cos{\left[ \frac{\pi}{3}+\frac{4\pi}{3}\cos{\left(\frac{\pi}{6}+\theta \right)} \right]} \right) \\ & \small \qquad + \frac{1}{2} \left( \cos{\left[ \frac{\pi}{3}-\frac{4\pi}{3}\cos{\left(\frac{\pi}{6}-\theta \right)} \right]} - \cos{\left[ \frac{\pi}{3}+\frac{4\pi}{3}\cos{\left(\frac{\pi}{6}+\theta \right)} \right]} \right) \\ & \text{Define the second term of RHS as a separate function and simplify by sum-to-product:} \\ & \color{red}{g(\theta)} = \small \cos{\left[ \frac{\pi}{3}-\frac{4\pi}{3}\cos{\left(\frac{\pi}{6}-\theta \right)} \right]} - \cos{\left[ \frac{\pi}{3}+\frac{4\pi}{3}\cos{\left(\frac{\pi}{6}+\theta \right)} \right]} = 2 \sin{\left[ \frac{\pi}{3}-\frac{2\pi}{3}\sin{\left( \theta \right)} \right]} \sin{\left[ \frac{2\pi}{\sqrt{3}}\cos{\left( \theta \right)} \right]} \\ & \text{And simplify the first term of RHS by sum-to-product to get:} \\ & \small \cos{\left[ \frac{2\pi}{3}-\frac{4\pi}{3}\sin{\theta} \right]} + \cos{\left[ \frac{\pi}{3}+\frac{4\pi}{3}\cos{\left(\frac{\pi}{6}+\theta \right)} \right]} \normalsize = g(\theta-2\pi/3) \Rightarrow \\ & f(\theta)=g(\theta-2\pi/3)+g(\theta)/2 \Rightarrow\quad \color{red}{-2f(\theta-\frac{\pi}{2})=g(\frac{\pi}{6}-\theta)+g(\frac{\pi}{6}+\theta)} \small\quad\text{ "even&symmetric"} \\ & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \\ \end{align} $$

Calculate $g(\theta)$ zeros by using its product formula, set $g(\pi/6-\theta)=g(\pi/6+\theta)$ to calculate $f(\theta-\pi/2)$ zeros, and reverse the substitutions $\{ \theta \rightarrow t \rightarrow x \}$ considering the definition ranges. The rest is yours.

$$ \begin{align} & \small g(\theta) = 2 \sin{\left[ \frac{\pi}{3}-\frac{2\pi}{3}\sin{\theta} \right]} \sin{\left[ \frac{2\pi}{\sqrt{3}}\cos{\theta} \right]} \Rightarrow \sin{\left[ \frac{\pi}{3}-\frac{2\pi}{3}\sin{\theta} \right]} = 0 \space\space\space\text{or}\space\space\space \sin{\left[ \frac{2\pi}{\sqrt{3}}\cos{\theta} \right]} = 0 \\ & \small \text{arcsin and solve} \space\Rightarrow \theta = \frac{n\pi}{3} = 0 \,, \pm\frac{\pi}{3} \,, \pm\frac{2\pi}{3} \,, \pm\pi \,, \text{...} \\ & \small + g(\frac{\pi}{6}-\theta) = - g(\frac{\pi}{6}+\theta) \Rightarrow \text{mirror and rotate } (\varphi \rightarrow -\varphi \rightarrow -\varphi-\pi) \space\Rightarrow + g(\frac{7\pi}{6}-\theta) = - g(\frac{7\pi}{6}+\theta) \\ & \small + 2 \sin{\left[ \frac{\pi}{3}-\frac{2\pi}{3}\sin{(\frac{\pi}{6}-\theta)} \right]} \sin{\left[ \frac{2\pi}{\sqrt{3}}\cos{(\frac{\pi}{6}-\theta)} \right]} = - 2 \sin{\left[ \frac{\pi}{3}-\frac{2\pi}{3}\sin{(\frac{\pi}{6}+\theta)} \right]} \sin{\left[ \frac{2\pi}{\sqrt{3}}\cos{(\frac{\pi}{6}+\theta)} \right]} \Rightarrow \\ & \small - 2 \sin{\left[ \frac{\pi}{3}+\frac{2\pi}{3}\sin{(\theta-\frac{\pi}{6})} \right]} \sin{\left[ \frac{2\pi}{\sqrt{3}}\cos{(\theta-\frac{\pi}{6})} \right]} = + 2 \sin{\left[ \frac{\pi}{3}+\frac{2\pi}{3}\sin{(\theta+\frac{\pi}{6})} \right]} \sin{\left[ \frac{2\pi}{\sqrt{3}}\cos{(\theta+\frac{\pi}{6})} \right]} \Rightarrow \\ & \small - 2 \sin{\left[ \frac{\pi}{3}-\frac{2\pi}{3}\sin{(\theta-\frac{7\pi}{6})} \right]} \sin{\left[ \frac{2\pi}{\sqrt{3}}\cos{(\theta-\frac{7\pi}{6})} \right]} = + 2 \sin{\left[ \frac{\pi}{3}-\frac{2\pi}{3}\sin{(\theta+\frac{7\pi}{6})} \right]} \sin{\left[ \frac{2\pi}{\sqrt{3}}\cos{(\theta+\frac{7\pi}{6})} \right]} \\ & \\ & -2f(\theta-\frac{\pi}{2})=g(\frac{\pi}{6}-\theta)+g(\frac{\pi}{6}+\theta) \Rightarrow -2f(\frac{\pi}{2}-\theta)=g(\frac{7\pi}{6}-\theta)+g(\frac{7\pi}{6}+\theta) \\ & f(\theta-\frac{\pi}{2})=0 \small \Rightarrow g(\frac{\pi}{6}-\theta) = - g(\frac{\pi}{6}+\theta) \Rightarrow g(\frac{7\pi}{6}-\theta) = - g(\frac{7\pi}{6}+\theta) \Rightarrow \normalsize \color{red}{f(\frac{\pi}{2}-\theta)=0} \\ & \text{Thus, No zeros for } f(\theta-\frac{\pi}{2}) \text{ in the range } ]0 \space\,, \frac{\pi}{3}[ \text{ because both } \small \space\space g(\frac{\pi}{6}-\theta) \,,\space \space\space g(\frac{\pi}{6}+\theta) \normalsize \lt 0 \\ & \text{Also, No zeros for } f(\frac{\pi}{2}-\theta) \text{ in the range } ]\frac{2\pi}{3} \,, \pi[ \text{ because both } \small g(\frac{7\pi}{6}-\theta) \,,\space g(\frac{7\pi}{6}+\theta) \normalsize \lt 0 \\ & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \\ \end{align} $$

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.

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 $0\lt x\le \dfrac{\pi}{3}$.