Question

4cos(x)²+8cos(x)+3=0

Alguém pfv resolve isso

Answer 1

$\displaystyle \sf 4cos^2(x)+8cos(x)+3=0 \\\\ \underline{Fa{\c c}amos}} : \\\\ cos(x) = a \\\\ \underline{Da{\'i}}} : \\\\ 4a^2+8a+3=0\\\\ a = \frac{-8\pm\sqrt{8^2-4\cdot 4\cdot 3 }}{2\cdot 4} \to a = \frac{-8\pm\sqrt{64-48}}{8} \\\\\\ a=\frac{-8\pm\sqrt{16}}{8} \to a = \frac{-8\pm4}{8} \\\\\\\ \left\{ \begin{array}{I} \displaystyle \sf a = \frac{-8-4}{8} \to a = \frac{-12}{8} \to a = \frac{-3}{2} \\\\ \displaystyle \sf a=\frac{-8+4}{8} \to a=\frac{-4}{8} \to a = \frac{-1}{2} \end{array} \right$

Desfazendo a troca de variável :

$\displaystyle \sf cos(x) = \frac{-3}{2} \to cos(x) = -1,5 \to (SEM \ SOLU{\c C}{\~A}O ) \\\\\\ cos(x) = \frac{-1}{2} \to cos(x) = cos\left(\frac{2\pi}{3}+2\cdot k\cdot \pi\right)=cos\left(\frac{4\pi}{3} +2\cdot k\cdot \pi \right)$

Portanto as soluções são :

$\displaystyle \boxed{\sf x = \left(\frac{2\pi}{3}+2\cdot k\cdot \pi\right) \ \ ;\ k\in\mathbb{Z} \ }\checkmark \\\\\\ \sf OU \\\\\ \boxed{\sf x = \left(\frac{4\pi}{3}+2\cdot k\cdot \pi \right)\ \ ;\ k\in\mathbb{Z} \ }\checkmark$