Question

Alguém consegui resolver:
2 sin^{2} x - 3 sin x + 1 = 0

HELP!!!!!

Answer 1

$2\text{sen}^2(\text x) - 3\text{sen(x)}+1 = 0$

é uma equação do 2º grau na variável sen(x). Usando bhaskpara temos :

$\displaystyle \text{sen(x)} = \frac{-(-3)\pm\sqrt{(-3)^2-4.2.1}}{2.2} \to \text{sen(x)} = \frac{3\pm\sqrt{9-8}}{4}$

$\displaystyle \text{sen(x)} = \frac{3\pm\sqrt{9-8}}{4} \to \text{sen(x)} = \frac{3\pm1}{4}$

então :

$\displaystyle \text{sen(x)} =\frac{3+1}{4} \to \boxed{\text{sen(x)} = 1}$

$\displaystyle \text{sen(x)} = 1 \to \boxed{\text x = \frac{\pi}{2} + 2.k.\pi} \ , \boxed{(\text k \in \mathbb{Z})}$

ou

$\displaystyle \text{sen(x)} =\frac{3-1}{4} \to \boxed{\text{sen(x)} = \frac{1}{2}}$

$\displaystyle \text{sen(x)} = \frac{1}{2} \to \boxed{\text x = \frac{\pi}{6} +2.\text k. \pi} \ , \ \boxed{(\text k \in \mathbb{Z})}$

OU

$\displaystyle \text{sen(x)} = \frac{1}{2} \to \boxed{\text x = \frac{5.\pi}{6} +2.\text k. \pi} \ , \ \boxed{(\text k \in \mathbb{Z})}$

Soluções :

$\displaystyle\boxed{\text x = \frac{\pi}{2} + 2.\text k.\pi} \ ; \ \boxed{\text x = \frac{\pi}{6} + 2.\text k.\pi} \ ; \ \boxed{\text x = \frac{5\pi}{6} + 2.\text k.\pi}}$, Com $\boxed{\text k \in \mathbb{Z}}$