Question

90 conjunto solução da equação 2cos2x+cosx-1=0, no universo
U=[0,2 tt ​

Answer 1

Temos :

$2\text{cos(2x)}+\text{cos(x)}-1=0$

Sabemos que :

$\text{cos(2x)}=2\text{cos}^2(\text x) - 1$

Substituindo :

$2.(2.\text{cos}^2(\text x) -1 ) + \text{cos(x)}-1=0$

$4.\text{cos}^2(\text x) -2 + \text{cos(x)}-1=0$

$4.\text{cos}^2(\text x) + \text{cos(x)}-3=0$

$\displaystyle \text{cos(x)} =\frac{-1\pm\sqrt{1-4.4.(-3)}}{2.4}\to \text{cos(x)}= \frac{-1\pm\sqrt{49}}{8}\\\\ \text{cos(x)} = \frac{-1\pm7}{8}\\\\ \text{cos(x)} = -1 \ ; \ \text{cos(x)} =\frac{3}{4}$

temos :

$\text{cos(x)} = -1 \\\\\boxed{\text x = \pi +2\text k.\pi \ , \ \text{com k} \in \mathbb{Z}}$

$\displaystyle \text{cos(x)} = \frac{3}{4} \\\\ \boxed{\text x = \text{ArcCos}(\frac{3}{4})+2.\text k.\pi \ , \ \text{com k} \in \mathbb{Z}}$

Conjunto solução :

$\huge\boxed{\ \text x = \pi +2.\text k.\pi \ , \ \text{com k} \in \mathbb{Z} }\checkmark$

$\huge\boxed{\ \text x = \text{ArcCos}(\frac{3}{4}) +2.\text k.\pi \ , \ \text{com k} \in \mathbb{Z} }\checkmark$

Obs : Não entendi esse conjunto universo.

Se for o intervalo for de $[0, 2\pi]$ então k = 0.