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