Question

Решите уравнение:

2sin^2⁡х + 3sin⁡х - 2 = 0

Найдите корни уравнения:

3sin^2⁡х - 7sin⁡х cos⁡х + 2cos^2⁡х = 0​

Answer 1

Ответ:

$2 { \sin(x) }^{2} + 3 \sin(x) - 2 = 0 \\ \sin(x) = y \\ 2 {y}^{2} + 3y - 2 = 0 \\ d = 9 + 4 \times 2 \times 2 = 25 \\ y = \frac{ - 3 - 5}{4} = - 2 \\ y = \frac{ - 3 + 5}{4} = \frac{1}{2} \\ \sin(x) = - 2 \\ x = arc \sin( - 2) + \pi \: n \\ \sin(x) = - \frac{1}{2} \\ x = \frac{7\pi}{6} + \pi \: n \\ \\ 3 { \sin(x) }^{2} - 7 \sin(x \cos(x) ) + 2 { \cos(x) }^{2} = 0 \: \: \: \: \div { \cos(x) }^{2} \\ 3 { \tan(x) }^{2} - 7 \tan(x) + 2 = 0 \\ \tan(x) = y \\ 3 {y}^{2} - 7y + 2 = 0 \\ d = 49 - 4 \times 3 \times 2 = 25 \\ y = \frac{7 - 5}{6} = \frac{1}{3} \\ y = \frac{7 + 5}{6} = 2 \\ x = arc \tan( \frac{1}{3} ) + \pi \: n \\ x = arc \tan(2) + \pi \: n$