Если не трудно решите))буду очень благодарен:3

Решите задачу:

$cos\frac{3\pi+x}{3}*cos\frac{9\pi+2x}{6}=-\frac{5}{48}*tg(2arctg1.5)\\ cos\frac{6\pi+2x}{6}*cos\frac{9\pi+2x}{6} = -\frac{5}{48}*\frac{2tg(arctg1.5)}{1-tg^2(arctg1.5)} \\ -0.5sin\frac{2x}{3} = -\frac{5}{48}* \frac{3}{1-1.5^2}\\ -0.5sin\frac{2x}{3} = -\frac{5}{48}*\frac{3}{-\frac{5}{4}}\\ -0.5sin\frac{2x}{3} = \frac{5*12}{48*5}\\ sin\frac{2x}{3}= -\frac{1}{2}\\ x=-\frac{\pi}{4} +3\pi*n\\ x = \frac{7\pi}{4}+3\pi*n$