Question

sin^2(3x)=3cos^2(3x)

Answer 1

$\frac{1 - \cos(6x) }{2} = \frac{3(1 + \cos(6x)) }{2}$
$3 + 3 \cos(6x) - 1 + \cos(6x) = 0$
$4 \cos(6x) = - 2$
$\cos(6x) = - \frac{1}{2}$
$6x = + - \frac{2}{3} \pi + 2\pi \times k$
$x = + - \frac{\pi}{9} + \frac{\pi \times k}{3}$