Question

Resolva a equação sen2 x+ 4cos x = - 4,
com 0 ≤ x ≤ 2π e assinale a alternativa com a resposta correta:

A) 0
B) π
C) /2
D) 3/2
E) 2π

Answer 1

$\displaystyle \sf sen^2x+4cos(x)=-4\\\\ \underline{\text{usando a rela{\c c}{\~a}o fundamental }}: \\\\ sen^2x+cos^2(x)=1 \\\\ sen^2(x)=1-cos^2(x) \\\\ \underline{Da{\'i}}}: \\\\ sen^2(x)+4cos(x)=-4 \\\\ 1-cos^2(x)+4cos(x) = -4 \\\\ -cos^2(x)+4cos(x)+5 = 0 \ \ \ \cdot (-1) \\\\ cos^2(x)-4cos(x)-5 = 0 \\\\ cos^2(x)-4cos(x) -5+9 = 9 \\\\ cos^2(x) -4cos(x)+4=9 \\\\\ \left(cos(x)-2\right)^2=9 \\\\\ cos(x)-2 = \pm 3 \\\\ cos(x) = 3+2 \to cos(x) = 5 \ (\text{N{\~a}o conv{\'e}m)}$

$\displaystyle \sf cos(x) = -3+2 \to cos(x) = -1 \to \huge\boxed{\sf x = \pi }\checkmark$

letra B