Question

Ребята решите 5 задание прошу!!! Даю 60 баллов!!!!!

Answer 1

3Sin²α - 4Sin2α = 3Cos²α

- 4Sin2α = 3Cos²α - 3Sin²α

- 4Sin2α = 3(Cos²α - Sin²α)

- 4Sin2α = 3Cos2α

Разделим обе части на Cos2α ≠ 0

- 4tg2α = 3

$tg2\alpha=-\frac{3}{4}\\\\tg2\alpha=\frac{2tg\alpha }{1-tg^{2}\alpha}\\\\\frac{2tg\alpha }{1-tg^{2}\alpha}=-\frac{3}{4}\\\\8tg\alpha=-3+3tg^{2}\alpha\\\\3tg^{2}\alpha -8tg\alpha-3=0\\\\\frac{\pi }{2}< \alpha <\pi \Rightarrow tg\alpha=-\frac{1}{3}\\\\Sin2\alpha=\frac{2tg\alpha}{1+tg^{2}\alpha} =\frac{2*(-\frac{1}{3})}{1+(-\frac{1}{3})^{2}}=\frac{-\frac{2}{3}}{\frac{10}{9}}=-\frac{3}{5}\\\\15Sin2\alpha=15*(-\frac{3}{5})=-9$