- [tex]\cos(2x)=\cos^2x-\sin^2x[/tex]
- [tex]\cos^2x+\sin^2(x)=1[/tex]
[tex]\cos(2x)+\sin x=0\\\\\cos^2x-\sin^2x+\sin x=0\\\\1-\sin^2x-\sin^2x+\sin x=0\\\\1 -2\sin^2x+\sin x=0\\\\[/tex]
A partir daqui vamos tomar [tex]\sin x = y[/tex]
[tex]1 -2\sin^2x+\sin x=0\\\\1-2y^2+y=0\\\\y=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\y'=\frac{-1+\sqrt{1^2-4(-2)*1}}{2(-2)}\\\\y'=\frac{-1+3}{-4}\\\\y'=-\frac{1}{2}\\\\\\y''=\frac{-1-3}{-4}\\\\y''=1[/tex]
Agora que encontramos os valores de y temos de voltar à variável original:
[tex]\sin x = y\\\\\sin x = \frac{-1}{2}\:\:ou\:\: \sin x=1[/tex]
1) Para a primeira possibilidade temos que lembrar que o seno possui uma simetria (veja figura)
[tex]\sin x = \frac{-1}{2}\\\\x'=\pi+\frac{\pi}{6}\\\\x'=\frac{7\pi}{6}+2k\pi\\\\\\x''=2\pi-\frac{\pi}{6}\\\\x''=\frac{11\pi}{6}+2k\pi[/tex]
2)
[tex]\sin x =1\\\\x''' = \frac{\pi}{2}+2k\pi[/tex]
A solução é, portanto:
[tex]S=\{x\in \mathbb{R}\: |\: x=\frac{\pi}{2}+\frac{2\pi k}{3},\:k \in \mathbb{Z} \}[/tex]
