Question

Если не трудно решите))буду очень благодарен:3

Answer 1

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