Question

знайдіть корені рівняння 6cos²x+5sinx-7=0​

Answer 1

Ответ:

$6 { \cos }^{2} (x )+ 5 \sin(x) - 7 = 0 \\ 6(1 - { \sin }^{2}(x)) + 5 \sin(x) - 7 = 0 \\ 6 - 6 \sin ^{2} (x ) + 5 \sin(x) - 7 = 0 \\ - 6 \sin {}^{2} (x ) + 5 \sin(x) - 1 = 0 \\ 6 \sin {}^{2} (x) - 5 \sin(x) - 1 = 0 \\ \sin(x) = y \\ 6 {y}^{2} - 5y - 1 = 0 \\ d = 25 + 24 = 49 \\ y1 = \frac{5 + 7}{12} = 1 \\ y2 = \frac{5 - 7}{12} = - \frac{1}{6}$

$\sin(x) = 1 \\ x = \frac{\pi}{2} + \pi \: n \\ \\ \sin(x) = - \frac{1}{6} \\ x = {( - 1)}^{n} arcsin( - \frac{1}{6} ) + \pi \: n$