Question

resolva equação:sen⁴x+cos⁴x=½?​

Answer 1

$\displaystyle \text{sen}^4(\text x)+\text{cos}^4(\text x)=\frac{1}{2 } \\\\\\ \underline{\text{Vamos fazer o seguinte}}: \\\\ \text{sen}^2(\text x)+\text{cos}^2(\text x) = 1 \\\\ \underline{\text{elevando ao quadrado}}: \\\\ ( \text{sen}^2(\text x)+\text{cos}^2(\text x) )^2=1^2 \\\\ \text{sen}^4(\text x)+\text{cos}^4(\text x)+2.\text{sen}^2(\text x).\text{cos}^2(\text x) = 1\\\\ \text{sen}^4(\text x)+\text{cos}^4(\text x) = 1-2.\text{sen}^2(\text x).\text{cos}^2(\text x)$

$\displaystyle \underline{\text{usando o arco dobro do seno}} : \\\\\ 2.\text{sen}(\text x)\text{cos}(\text x) = \text{sen}(2\text x) \\\\ 4.\text{sen}^2(\text x).\text{cos}^2(\text x)=\text{sen}^2(2\text x) \\\\ \text{sen}^2(\text x).\text{cos}^2(\text x) = \frac{\text{sen}^2(2\text x)}{4}$

Então :

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

substituindo na equação original :  

$\displaystyle 1-\frac{\text{sen}^2(2.\text x)}{2} = \frac{1}{2} \\\\\\ 2-\text{sen}^2(2.\text x)=1 \\\\\\ \text{sen}^2(2.\text x) = 1 \\\\\\ \text{sen}(2.\text x) = \pm 1 \\\\ \underline{\text{ent{\~a}o}}: \\\\ \text{sen}(2.\text x) = 1 \to 2\text x = \frac{\pi}{2}+ 2.\text k.\pi \to \text x= \frac{\pi}{4} + \text k.\pi \ , \ \text k \in \mathbb{Z} \\\\\\ \text{sen}(2.\text x) = -1 \to 2\text x = \frac{3\pi}{2}+ 2.\text k.\pi \to \text x= \frac{3\pi}{4} + \text k.\pi \ , \ \text k \in \mathbb{Z}$

Soluções :

$\huge\boxed{\ \text x = \frac{\pi}{4}+ \text k.\pi \ \ , \ \ \text k \in \mathbb{Z}\ } \checkmark \\\\\\ \text{ou} \\\\\\ \huge\boxed{\ \text x = \frac{3\pi}{4}+\text k.\pi \ \ , \ \text k \in \mathbb{Z} \ }\checkmark$