Question

Решите пожалуйста

2sin^2x+3sinx+1=0

Answer 1

Ответ:

$\boxed{ \left[ \begin{gathered} x = (-1)^{n + 1} \left \dfrac{\pi}{6} + \pi n \\ \ x = -\dfrac{\pi}{2} + 2\pi n \end{gathered} \ \ \ n \in \mathbb Z }$

Объяснение:

$2 \sin^{2} x + 3 \sin x + 1 = 0$

Замена: $\sin x = t; t \in [-1;1]$

$2t^{2} + 3t + 1 = 0$

$D = 9 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 = 1^{2}$

$t_{1} = \dfrac{-3 + 1}{2 \cdot 2} = \dfrac{-2}{4} = -\dfrac{1}{2}$

$t_{2} = \dfrac{-3 - 1}{2 \cdot 2} = \dfrac{-4}{4} = -1$

$\left[ \begin{gathered} \sin x = -\dfrac{1}{2} \\ \sin x = -1 \end{gathered}$  $\left[ \begin{gathered} x = (-1)^{n} arcsin \left ( -\dfrac{1}{2} \right) + \pi n \\ \ x = -\dfrac{\pi}{2} + 2\pi n \end{gathered} \ \ \ n \in \mathbb Z$

$\left[ \begin{gathered} x = (-1)^{n} \left ( -\dfrac{\pi}{6} \right) + \pi n \\ \ x = -\dfrac{\pi}{2} + 2\pi n \end{gathered} \ \ \ n \in \mathbb Z$     $\left[ \begin{gathered} x = (-1)^{n + 1} \left \dfrac{\pi}{6} + \pi n \\ \ x = -\dfrac{\pi}{2} + 2\pi n \end{gathered} \ \ \ n \in \mathbb Z$