2cos(2α) + 7sinα = 0чему равны sinα и cos2α?

2Cos2α + 7Sinα = 0

2(1 -2Sin²α) + 7Sinα =0

2 - 4Sin²α + 7Sinα = 0

4Sin²α - 7Sinα -2 = 0

Сделаем замену :

Sinα = m , - 1 ≤ m ≤ 1

4m² - 7m - 2 = 0

D = (-7)² - 4 * 4 * (- 2) = 49 + 32 = 81 = 9²

1" alt="m_{1}=\frac{7-9}{8}=-\frac{2}{8}=-\frac{1}{4}\\\\m_{2}=\frac{7+9}{8}=2>1" align="absmiddle" class="latex-formula">

m = 2 - посторонний корень

$Sin\alpha=-\frac{1}{4}\\\\Cos2\alpha=1-2Sin^{2}\alpha=1-2*(-\frac{1}{4})^{2}=1-2*\frac{1}{16}=1-\frac{1}{8}=\frac{7}{8}$