Question

4-3sin²x=5sinx
ПОМОГИТЕЕЕЕ!!!!!

Answer 1

Ответ:

$\boldsymbol{\boxed{x = (-)^{k}\arcsin \bigg ( \dfrac{\sqrt{73} - 5}{6} \bigg ) + \pi k, k \in \mathbb Z}}$

Объяснение:

$4 - 3 \sin^{2} x = 5 \sin x$

$- 3 \sin^{2} x - 5 \sin x + 4 = 0| \cdot (-1)$

$3 \sin^{2} x + 5 \sin x - 4 = 0$

Замена: $\sin x = t; t \in [-1;1]$

$3t^{2} + 5t - 4 = 0$

$D = 25 - 4 \cdot 3 \cdot (-4) = 25 + 48 = 73$

$\boxed{t_{1} = \dfrac{-5 + \sqrt{73} }{6} \in [-1;1]}$

$t_{2} = \dfrac{-5 - \sqrt{73} }{6} \notin [-1;1]$

$\sin x = \dfrac{\sqrt{73} - 5}{6}$

$x = (-)^{k}\arcsin \bigg ( \dfrac{\sqrt{73} - 5}{6} \bigg ) + \pi k, k \in \mathbb Z$