Question

8cos x $\left({\cos }^2\frac{\mathrm{\pi }}{6}-{\sin }^{2}x-\frac{1}{2}\right)=1$

$\Rightarrow$8 cos x $\left(\frac{1}{4}-(1-{\cos }^{2}x)\right)=1$
$\Rightarrow$8 cos x $\left({\cos }^{2}x-\frac{3}{4}\right)=1$
$\Rightarrow$ 2cos³x = 1 $\Rightarrow$cos³x = $\frac{1}{2}$
$\therefore 3x + 2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{3}, \mathrm{n} \in \mathrm{I}$
$\Rightarrow x = \frac{2n\mathrm{\pi } }{3}\pm \frac{\mathrm{\pi }}{9}$
In x $\in$ [0, $\mathrm{\pi }$] : x = $\frac{\mathrm{\pi }}{9}, \frac{{\displaystyle 2\mathrm{\pi }}}{{\displaystyle 3}}+\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 9}}, \frac{{\displaystyle 2\mathrm{\pi }}}{{\displaystyle 3}}-\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 9}}$ only
sum = $\frac{13\mathrm{\pi }}{9}$