Question

Galera me ajudem na questão de trigonometria? Obrigado

Answer 1

$4\cos^3 x - \cos x \ \textless \ 0\\ \\ \cos x(4\cos^2 x -1)\ \textless \ 0\\ \\ \cos x (2\cos x - 1)(2\cos x + 1)\ \textless \ 0\\ \\ \cos x \in (-\infty;-\frac{1}{2}) \cup (0;\frac{1}{2}) \\ \\ \text{Pero... }|\cos x|\leq 1 \text{ entonces}\\ \\ \boxed{\cos x \in [-1;-\frac{1}{2}) \cup (0;\frac{1}{2})}$

Si $\cos \in[-1,-\frac{1}{2})$ entonces

                 $x\in\left(\frac{2\pi}{3},\frac{4\pi}{3}\right)$

Si $\cos x\in\left(0; \frac{1}{2}\right)$ entonces

                  $x\in\left(\frac{\pi}{3}:\frac{\pi}{2}\right)\cup \left(\frac{3\pi}{2};\frac{5\pi}{3}\right)$

Luego unimos

$\displaystyle x\in\left(\frac{2\pi}{3},\frac{4\pi}{3}\right) \cup \left(\frac{\pi}{3}:\frac{\pi}{2}\right)\cup \left(\frac{3\pi}{2};\frac{5\pi}{3}\right) \\ \\ \\ \boxed{x\in \left(\frac{\pi}{3}:\frac{\pi}{2}\right)\cup \left(\frac{2\pi}{3},\frac{4\pi}{3}\right) \cup \left(\frac{3\pi}{2};\frac{5\pi}{3}\right)}$