Question

2sin^2 x+sinx=1 pomogiteee​

Answer 1

$2 \sin ^{2} x + \sin x = 1\\2 \sin ^{2} x + \sin x - 1 = 0$

Сделаем замену: $\sin x = t, \ t \in [-1; 1]$

$2t^{2} + t - 1 = 0\\D = 1^{2} - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9\\t_{1} = \dfrac{-1 + \sqrt{9} }{2 \cdot 2} = \dfrac{-1 + 3}{4} = \dfrac{1}{2} \\t_{2} = \dfrac{-1 - \sqrt{9} }{2 \cdot 2} = \dfrac{-1 - 3}{4} = -1$

Обратная замена:

$\left[\begin{array}{ccc} \sin x = \dfrac{1}{2} \ \ \\ \sin x = -1\end{array}\right$

$\left[\begin{array}{ccc} x = (-1)^{n}\cdot \dfrac{\pi}{6} + \pi n, \ n \in Z \ \ \\ x = -\dfrac{\pi}{2} + 2\pi k, k \in Z \ \ \ \ \ \ \ \ \ \ \end{array}\right$

Ответ: $x = (-1)^{n}\cdot \dfrac{\pi}{6} + \pi n, \ n \in Z$, $x = -\dfrac{\pi}{2} + 2\pi k, k \in Z$