Question

Resolvendo, no intervalo [0, 2π[, a equação 2 sen²x – sen x = 1, a soma das raízes é:

Answer 1

$\sin(x)=u\\\\\\2u^2-u=1\\\\2u^2-u-1=0\\\\u'=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{(-1)^2-4.2(-1)}}{2.2}=\frac{1+\sqrt{9}}{4}=\frac{4}{4}=1\\\\u''=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{(-1)^2-4.2(-1)}}{2.2}=\frac{1-\sqrt{9}}{4}=\frac{-2}{4}=\frac{-1}{2}$

$\sin(x)=1\\\\x=\arcsin(1)\\\\x=\frac{\pi}{2}$

$\sin(x)=\frac{-1}{2}\\\\x=\arcsin(\frac{-1}{2})\\\\x=\frac{7\pi}{6}\:\:ou\:\:x=\frac{11\pi}{6}$

$\frac{\pi}{2}+\frac{7\pi}{6}+\frac{11\pi}{6}\\\\\frac{3\pi+7\pi+11\pi}{6}\\\\\frac{21\pi}{6}$