EDIT:

Actually major brain fart on the line where I apply 3^0. Better redo this.

EDIT 2:

Redid the problem. Hope it is right this time:

$\displaystyle log_{3}((2cos(x))(cos(x)))\geq 0$

$\displaystyle log_{3}(2cos^{2}(x)+cos(x))\geq 0$

$\displaystyle 3^{(2cos^{2}(x)+cos(x))}\geq 0$

$\displaystyle 3\geq (2cos^{2}(x)+cos(x))$

$\displaystyle 0\geq 2cos^{2}(x)+cos(x)-3$

$\displaystyle 0\geq (2cos(x)-3)(cos(x)+1)$

------------------------------------

$\displaystyle 0\geq (2cos(x)-13)$

$\displaystyle \frac{3}{2}\geq cos(x)$

All values of x are negative

OR

$\displaystyle -1\geq cos(x)$

All values of x are positive except x = pi

And since a positive x negative equals it will always be negative (or zero) and therefore all values of x in the given interval resolve the equation