Question

Resolva as seguintes equações trigonométricas:
(a) sec 2x = 2
(b) 2sen2x = senx

(c) cossec2x = −
√
2
(d) 2senx cos x =
√
2
2

Segue anexo:

Answer 1

$\large\boxed{\begin{array}{l}\sf a)~\rm sec(2x)=2\\\rm\dfrac{1}{cos(2x)}=2\longrightarrow cos(2x)=\dfrac{1}{2}\\\\\rm cos(2x)=cos\bigg(\dfrac{\pi}{3}\bigg)\\\\\rm 2x=\dfrac{\pi}{3}\\\\\rm x=\dfrac{\pi}{6}+2k\pi,k\in\mathbb{Z}\\\\\rm cos(2x)=cos\bigg(\dfrac{5\pi}{3}\bigg)\end{array}}$

$\large\boxed{\begin{array}{l}\rm 2x=\dfrac{5\pi}{3}\\\\\rm 6x=5\pi\\\\\rm x=\dfrac{5\pi}{6}+2k\pi,k\in\mathbb{Z}\end{array}}$

$\large\boxed{\begin{array}{l}\rm S=\bigg\{\dfrac{\pi}{6}+2k\pi,k\in\mathbb{Z}~ou~\dfrac{5\pi}{6}+2k\pi,k\in\mathbb{Z}\bigg\}\end{array}}$

$\large\boxed{\begin{array}{l}\sf b)~\rm 2sen^2(x)=sen(x)\\\rm 2sen^2(x)-sen(x)=0\\\rm sen(x)(2sen(x)-1)=0\\\rm sen(x)=0\\\rm x=2k\pi,k\in\mathbb{Z}\\\rm x=\pi+2k\pi,k\in\mathbb{Z}\\\rm 2sen(x)-1=0\\\rm 2sen(x)=1\\\rm sen(x)=\dfrac{1}{2}\\\\\rm x=\dfrac{\pi}{6}+2k\pi,k\in\mathbb{Z}\\\\\rm x=\dfrac{5\pi}{6}+2k\pi,k\in\mathbb{Z}\end{array}}$

$\large\boxed{\begin{array}{l}\rm S=\bigg\{2k\pi,k\in\mathbb{Z};\pi+2k\pi,k\in\mathbb{Z};\dfrac{\pi}{6}+2k\pi,k\in\mathbb{Z}\\\rm\dfrac{5\pi}{6}+2k\pi,k\in\mathbb{Z} \bigg\}\end{array}}$

$\large\boxed{\begin{array}{l}\sf c)~\rm cossec(2x)=-\sqrt{2}\\\rm\dfrac{1}{sen(2x)}=-\sqrt{2}\\\\\rm sen(2x)=-\dfrac{1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\\\\\rm 2x=\dfrac{5\pi}{4}+2k\pi\longrightarrow x=\dfrac{5\pi}{8}+k\pi,k\in\mathbb{Z}\\\\\rm 2x=\dfrac{7\pi}{4}+2k\pi\longrightarrow x=\dfrac{7\pi}{8}+k\pi,k\in\mathbb{Z}\\\\\rm S=\bigg\{\dfrac{5\pi}{8}+k\pi,k\in\mathbb{Z};\dfrac{7\pi}{8}+k\pi,k\in\mathbb{Z}\bigg\}\end{array}}$

$\large\boxed{\begin{array}{l}\sf d)~\rm 2sen(x)cos(x)=\dfrac{\sqrt{2}}{2}\\\\\rm sen(2x)=\dfrac{\sqrt{2}}{2}\\\\\rm 2x=\dfrac{\pi}{4}+2k\pi\longrightarrow x=\dfrac{\pi}{8}+k\pi,k\in\mathbb{Z}\\\\\rm 2x=\dfrac{3\pi}{4}+2k\pi\longrightarrow x=\dfrac{3\pi}{8}+k\pi,k\in\mathbb{Z}\\\\\rm S=\bigg\{\dfrac{\pi}{8}+k\pi,k\in\mathbb{Z};\dfrac{3\pi}{8}+k\pi,k\in\mathbb{Z}\bigg\}\end{array}}$