Question

Решите тригонометрическое уравнение​

Answer 1

sin x * (2 sin² x - 1) + cos² 2x = 0

cos 2x = 1 - 2 sin² x

-sin x * ( 1 - 2 sin² x ) + cos ² 2x = 0

- sin x * cos 2x + cos² 2x = 0 | : cos 2x (cos 2x ≠ 0)

cos 2x * ( - sin x + cos 2x) = 0

cos 2x * ( cos 2x - sin x ) = 0

cos 2x = 0

$2x = \frac{\pi }{2} + \pi n$ , n∈Z | : 2

$x = \frac{\pi }{4} + \frac{\pi n}{2}$

cos 2x - sin x  = 0

1 - 2 sin² x - sin x  = 0

- 2 sin² x - sin x + 1 = 0 | : (-1)

2 sin² x + sin x - 1 = 0

Пусть sin x = t, тогда

2t² + t - 1 = 0

D = 1² - 4 * 2 * (-1) = 1 + 8 = 9 = 3²

$t_{1} = \frac{-1+3}{2*2} = \frac{2}{4} = \frac{1}{2}$

$t_{2} = \frac{-1-3}{2*2} = - \frac{4}{4} = -1$

Вернёмся к замене:

sin x = 1/2

$x = (-1)^{n} * arcsin \frac{1}{2} + \pi k$, k∈Z  

$x = (-1)^{n} * \frac{\pi }{6} + \pi k$ , k∈Z

sin x = -1

$x = - \frac{\pi }{2} + 2\pi k$ , k∈Z

ответ: $x = \frac{\pi }{4} + \frac{\pi n}{2}$; $x = (-1)^{n} * \frac{\pi }{6} + \pi k$; $x = - \frac{\pi }{2} + 2\pi k$ , n∈Z  ,k∈Z