Solve trigonometric equation $\cos \sqrt{\frac{\pi^2-x^2}3} + 2\cos \frac{\pi -x}3=0$

I don't seem to have a credible way to solve the trigonometric equation below

$$\cos \sqrt{\frac{\pi^2-x^2}3} + 2\cos \frac{\pi -x}3=0$$

I was able to guess quickly that $-\pi$ is a root, and, after staring at it long enough, was able to see that $-\frac\pi2$ is another one. I graphed the function and verified that these two are the only roots. However, I have not been able to solve it algebraically.

One thing I have tried is to convert it into a system of equations $$3y^2=\pi^2 - x^2$$ $$\cos y +2\cos\frac{\pi-x}3=0$$ and hoped to explore the symmetry, as seen the plot

Edit:

Note that the plot reveals that one root is at an extreme, where the two curves are tangential to each other. This indicates that the root may be solved via calculus, albeit not algebraically. One easier example of such soluble equation is $\pi\cos x=\sqrt{\pi^2 -x^2}$ which has a single root at its extreme. The complication in the posted equation is that it has two roots.

Solution 1:

Let's set $$f:[-\pi,\pi]\to\Bbb R:f(x)=\cos \sqrt{\frac{\pi^2-x^2}3} + 2\cos \frac{\pi -x}3$$

The two cosine arguments are roughly equal in size, so it actually turns out that the scalar $2$ is very useful. Indeed, for every $x\in(-\pi,\pi]$, we have that $$\bigg|2\cos \frac{\pi -x}3\bigg|>\bigg|\cos \sqrt{\frac{\pi^2-x^2}3}\bigg|\implies \operatorname{sgn}(f(x))=\operatorname{sgn}(\cos\bigg(\frac{\pi-x}{3}\bigg))$$ where $\operatorname{sgn}(x)$ is the sign function.

Our task of proving that $-\pi$ and $-\frac \pi2$ are the only solutions is then reduced to showing that $\cos(\frac{\pi-x}{3})$ is strictly non-zero across $(-\pi,-\frac\pi 2)\cup (-\frac\pi 2, \pi)$ -n.b. it is negative on the first half and positive on the second half.