Question

Solve 2sin2x-sqrt (2) =0 in the interval [0, 2pi)

Answer 1

\[
2 \sin(2x) = \sqrt 2\\
\sin(x)= \frac{\sqrt 2}2

\]

Answer 2

What angles in \[ [0, 2\pi) \] have sin equal to \[ \frac{ \sqrt2} 2\]?

Answer 3

$$
2 \sin(2x) = \sqrt 2 \implies
\sin(2x)= \cfrac{\sqrt 2}{2}\\
sin^{-1}(\sin(2x)) = sin^{-1}\pmatrix{\cfrac{\sqrt 2}{2}}\\
2x= \color{red}{\alpha} \implies x=\cfrac{\color{red}{\alpha}}{2}
$$
as @eliassaab said, what are those angles?

Answer 4

\[
\frac \pi 4 \\
\frac { 3 \pi} 4

\]

Answer 5

Sory, I thought you were asking me?

Answer 6

hehe, :), anyhow, he's not around for now, he'll be back :)