item a)
[tex]\displaystyle \text{Sen}(\frac{\text x}{2})= 1[/tex]
Sabemos que :
[tex]\displaystyle \text{Sen}(\frac{\pi}{2}) = 1[/tex]
Porém a cada uma volta completa ([tex]2.\pi[/tex]) no círculo trigonométrico resultam no mesmo valor para o seno, logo :
[tex]\displaystyle \text{Sen}(\frac{\pi}{2}+2.\text k.\pi )= 1 \ , \boxed{\text{com k} \in \mathbb{Z}}[/tex]
Então basta igualarmos, assim :
[tex]\displaystyle \text{Sen}(\frac{\text x}{2})= \text{Sen}(\frac{\pi}{2}+2.\text k.\pi )[/tex]
[tex]\displaystyle \frac{\text x}{2} = \frac{\pi }{2} +2.\text k.\pi \ , \ (\text{com k }\in \mathb{Z})[/tex]
[tex]\displaystyle \frac{\text x}{2} = \frac{\pi + 4.\text k.\pi}{2} \ , \ (\text{com k }\in \mathb{Z})[/tex]
Multiplicando os dos lados por dois, temos que :
[tex]\huge\boxed{\displaystyle \text x = \pi +4.\text k.\pi \ , \ (\text{com k }\in \mathbb{Z})} \checkmark[/tex]
item b)
[tex]\text{Sen(x)}^2 -2.\text{Sen(x)}+1=0[/tex]
Fazendo y = Sen(x) :
[tex]\text y^2-2\text y +1=0[/tex]
[tex](\text y-1)^2 = 0 \to \text y -1 = 0[/tex]
[tex]\text y = 1[/tex]
então :
[tex]\text{Sen(x)} = 1[/tex]
Usando a mesma ideia do item (a), temos :
[tex]\displaystyle \text{Sen(x)} =\text{Sen}(\frac{\pi}{2}+2.\text k.\pi )[/tex]
[tex]\huge\boxed{\displaystyle \text x = \frac{\pi}{2} +2.\text k.\pi \ ,\ (\text{com k } \in \mathbb{Z}) }\checkmark[/tex]
