Question

Решите:
4cos²x-3sinx=3
Пжж дам 50 баллов

Answer 1

Ответ:

Ответ: -π/2 + 2πk, (-1)ⁿ·arcsin(1/4)+πn, где k,n - целые числа.

Answer 2

Ответ:

$\boxed{ \left[ \begin{gathered} \left\ x = (-1)^{k} \cdot \arcsin( 0,25) +\pi k,k \in \mathbb Z \right. \hfill \\ \left\ x = \dfrac{\pi }{2} + 2\pi n, n \in \mathbb Z \end{gathered} \right. }$$\left[ \begin{gathered} \left\ x = (-1)^{k} \cdot \arcsin( 0,25) +\pi k,k \in \mathbb Z \right. \hfill \\ \left\ x = \dfrac{\pi }{2} + 2\pi n, n \in \mathbb Z \end{gathered} \right.$

Объяснение:

$4\cos^{2} x - 3 \sin x = 3$

$4( 1 - \sin^{2} x) - 3 \sin x = 3$

$4 - 4\sin^{2} x - 3 \sin x = 3$

$4\sin^{2} x + 3 \sin x - 1 = 0$

Замена: $\sin x = t; t \in [-1;1]$

$4t^{2} + 3t - 1 = 0$

$D = 9 - 4 \cdot 4 \cdot (-1) = 9 + 16 = 25 = 5^{2}$

$t_{1} = \dfrac{-3 + 5}{2 \cdot 4} = \dfrac{2}{2 \cdot 4} = \dfrac{1}{4} =0,25$

$t_{2} = \dfrac{-3 - 5}{2 \cdot 4} = \dfrac{-8}{2 \cdot 4} = -\dfrac{8}{9} = -1$

$t_{1},t_{2} \in [-1;1]$

$\left[ \begin{gathered} \left\ \sin x = t_{1} \right. \hfill \\ \left\ \sin x = t_{2} \end{gathered} \right.$  $\left[ \begin{gathered} \left\ \sin x = 0,25 \right. \hfill \\ \left\ \sin x = -1 \end{gathered} \right.$  $\left[ \begin{gathered} \left\ x = (-1)^{k} \cdot \arcsin( 0,25) +\pi k,k \in \mathbb Z \right. \hfill \\ \left\ x = \dfrac{\pi }{2} + 2\pi n, n \in \mathbb Z \end{gathered} \right.$