Question

URGENTE:
Existem dois valores reais de x, com 0 ⩽ x < 2π, tais que 2cosˆ2 x + 9senx = 6.


Qual é o maior desses valores de x?


A- π/3

B- π/2

C- 2π/3

D- 5π/6

E π

Answer 1

$2\cos^2x+9\sin x=6\\\\2(1-\sin^2x)+9\sin x=6\\\\-2\sin^2x+9\sin x-4=0\\\\2\sin2x-9\sin x+4=0\\\\2y^2-9y+4=0\\\\y'=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{9+\sqrt{81-32}}{4}=\frac{9+7}{4}=4\\\\y''=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{9-\sqrt{81-32}}{4}=\frac{9-7}{4}=\frac{1}{2}\\\\\sin x=4\:\:ou\:\:\sin x=\frac{1}{2}$

$\sin x =\frac{1}{2}\\\\\arcsin\frac{1}{2}=x\\\\x=\frac{\pi}{6}\:\:ou\:\:x=\frac{5\pi}{6}$

Portanto, D