Question

Resolva a equação 2 sen² x - sen x -1 = 0

Answer 1

$\displaystyle \sf 2 sen^2x-sen\ x-1 = 0 \\\\ \underline{\text{Fa{\c c}amos}} : \\\\ sen\ x = a \\\\ \underline{Da{\'i}}} : \\\\ 2a^2-a -1 = 0 \\\\ a = \frac{-(-1)\pm\sqrt{(-1)^2-4\cdot 2 \cdot (-1)}}{2\cdot 2} \\\\\\ a = \frac{1\pm\sqrt{1+8}}{4} \to a =\frac{1\pm\sqrt{9}}{4} \\\\\\ a = \frac{1\pm3}{4} \\\\ \left\{\begin{array}{I} \displaystyle \sf a=\frac{1+3}{4} \to a = 1 \\\\ \displaystyle \sf a = \frac{1-3}{4} \to a = \frac{-1}{2}\end{array} \right$

Desfazendo a troca de variável :

*

$\displaystyle \\\ \sf a = 1 \\\\ sen\ x = 1\\\\ x = \left(\frac{\pi }{2} +2\cdot k\cdot \pi\right) \ , \ k \in\mathbb{Z}$

*

$\displaystyle \\\ \sf a = \frac{-1}{2} \\\\ sen\ x = \frac{-1}{2}\\\\ x = \left(\frac{7\pi }{6} +2\cdot k\cdot \pi\right) \ , \ k \in\mathbb{Z} \\\\ ou \\\\ x = \left(\frac{11\pi }{6} +2\cdot k\cdot \pi\right) \ , \ k \in\mathbb{Z}$

Portanto as soluções são :

$\displaystyle \boxed{\sf x = \left(\frac{ \pi }{2} +2\cdot k\cdot \pi\right) \ , \ k \in\mathbb{Z}}\checkmark \\\\\\ \boxed{\sf x = \left(\frac{7\pi }{6} +2\cdot k\cdot \pi\right) \ , \ k \in\mathbb{Z}}\checkmark \\\\\\ \boxed{\sf x = \left(\frac{11\pi }{6} +2\cdot k\cdot \pi\right) \ , \ k \in\mathbb{Z} }\checkmark$