Where are these additional solutions coming from? Solve for $x$:  $2\sin(2x)-\sqrt{2} = 0$ in interval $[0,2\pi)$ 
Step $1$: Add $\sqrt{2}$ and divide by $2$ to get $\sin(2x) = \dfrac{\sqrt{2}}{2}$
Step $2$: Set $2x$ equal to the angles where $\sin(x) = \dfrac{\sqrt{2}}{2}$:
$2x = \dfrac{\pi}{4}$  and $2x = \dfrac{3\pi}{4}$
Step $3$: Solve for $x$ by dividing by $2$: $x = \dfrac{\pi}{8}$  and $x = \dfrac{3\pi}{8}$
My textbook also lists $\dfrac{9\pi}{8}$  and $\dfrac{11\pi}{8}$ as additional solutions, anyone know where they may have came from? thanks

$x\in[0,2\pi)$   means   $2x\in[0,4\pi)$, so you have find the solutions in this larger interval.
