Question

resolver, para 0 < × < 2pi, a equação 2 sen x + raiz de 2 = 0

Answer 1

$2\sin(x)+\sqrt2=0\\\\2\sin(x)=-\sqrt2\\\\\sin(x)=-\dfrac{\sqrt2}{2}$


Dois ângulos compreendidos entre $0$ e $2\pi$ têm o seno igual a $-\dfrac{\sqrt2}{2}$:

$\begin{matrix}\sin(x)=-\dfrac{\sqrt2}{2}\\\\\sin(x)=\sin\left(\dfrac{5\pi}{4}\right)\\\\\therefore\\\\x=\dfrac{5\pi}{4}\end{matrix}\qquad\qquad\qquad\qquad\begin{matrix}\sin(x)=-\dfrac{\sqrt2}{2}\\\\\sin (x)=\sin\left(\dfrac{7\pi}{4}\right)\\\\\therefore\\\\x=\dfrac{7\pi}{4}\end{matrix}\\\\\\\\\boxed{S=\left\{\dfrac{5\pi}{4}{,}\ \dfrac{7\pi}{4}\right\}}$