Question

resolva as equações, considerando U= [0,2 pí]
a) cos x = 1/2

b) na imagem
me ajudem plissss

Answer 1

$\displaystyle \text U=[\ 0, 2\pi \ ] \\\\ \text{item a)} \\\\ \text{cos x}=\frac{1}{2} \\\\\ \text{cos x}=\text{cos }(\frac{\pi}{3} + 2\text k\pi) \to \text x = \frac{\pi}{3}+2\text k.\pi \ , \text k\in \mathbb{Z}_{+} \\\\ \text{cos x}=\text{cos }(\frac{5\pi}{3}+2\text k.\pi )\to \text x = \frac{5\pi}{3} + 2\text k.\pi \ ,\ \text k \in \mathbb{Z}_{+}$

$\displaystyle \underline{\text{Portanto as solu{\c c}{\~o}es s{\~a}o}}: \\\\ \huge\boxed{\text x =\frac{\pi}{3}+2\text k.\pi \ , \ \text k \in \mathbb{Z}_{+}\ }\checkmark \\\\\\ \text{ou} \\\\\ \huge\boxed{\text x =\frac{5\pi}{3}+2\text k.\pi \ , \ \text k \in \mathbb{Z}_{+}\ }\checkmark$

item b)

$\displaystyle \text{12 cos}^2\text x-9=0 \\\\ \text{cos}^2\text x=\frac{9}{12} \to \text{cos}^2\text x= \frac{3}{4} \\\\\\ \text{cos x}=\pm \frac{\sqrt{3}}{\sqrt{4}} =\pm\frac{\sqrt{3}}{2} \\\\\\ 1) \\\\ \text{cos x}=\frac{\sqrt{3}}{2} \\\\\ \boxed{\text x = \frac{\pi}{6} +2\text k.\pi \ , \ \text k \in \mathbb{Z}_{+}} \ ; \ \boxed{\text x = \frac{11\pi}{6}+2\text k.\pi \ , \ \text k \in \mathbb{Z}_{+}}$

$\displaystyle 2) \\\\ \text{cos x}=\frac{-\sqrt{3}}{2} \\\\ \boxed{\text x = \frac{5\pi}{6}+2\text k.\pi\ ,\ \text k \in \mathbb{Z}_{+}} \ ; \ \boxed{\text x = \frac{7\pi}{6}+2\text k.\pi\ ,\ \text k \in \mathbb{Z}_{+}}$

Portanto as soluções são :

$\huge\boxed{\boxed{\text x = \frac{\pi}{6}+2\text k.\pi\ ,\ \text k \in \mathbb{Z}_{+}}}\checkmakr \\\\\\ \huge\boxed{\boxed{\text x = \frac{11\pi}{6}+2\text k.\pi\ ,\ \text k \in \mathbb{Z}_{+}}}\checkmark \\\\\\ \huge\boxed{\boxed{\text x = \frac{5\pi}{6}+2\text k.\pi\ ,\ \text k \in \mathbb{Z}_{+}}}\checkmark \\\\\\ \huge\boxed{\boxed{\text x = \frac{7\pi}{6}+2\text k.\pi\ ,\ \text k \in \mathbb{Z}_{+}}}\checkmark$

(Obs : k pertence aos inteiros não negativos porque o intervalo é fechado em 0, logo não convém soluções negativas e tem que ter o $2\text k.\pi$ porque o intervalo é aberto no $2\pi$)