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

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@Sivaram: oops, I think I started (re)editing the question shortly after you began your edit...and before you finished! –  amWhy May 12 '11 at 0:26
@Amy: No problem. I have rolled back your edit. –  user17762 May 12 '11 at 0:43

Your step two is correct except for a minor omission. More properly,

$2x = \displaystyle \frac{\pi}{4} + 2k\pi$, $k$ integer, and

$2x = \displaystyle \frac{3\pi}{4} + 2k\pi$, $k$ integer

Dividing these two expressions by 2 yield

$x = \displaystyle \frac{\pi}{8} + k\pi$, $k$ integer, and

$x = \displaystyle \frac{3\pi}{8} + k\pi$, $k$ integer

While $k=0$ gives the solutions you have, $k=1$ gives solutions that are ALSO in the interval $[0,2\pi]$. That's why your textbook has two additional solutions.

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

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How would you find these additional values, I just use the unit circle to find the first two. –  Matt May 12 '11 at 0:21
@Matt: 0 to $4\pi$ is two complete revolutions around the unit circle, so keep the two points you have, but get to them again by going around the unit circle once ($2\pi$), then to your solution points. –  Isaac May 12 '11 at 0:25
@Matt: you have to take the unit circle round another revolution so you also have the possibilities $2x = 9\pi / 4$ or $2x = 11\pi / 4$ –  Henry May 12 '11 at 0:31

$\sin(\theta) = \sqrt(2)/2$ iff $\theta = \pi/4 + 2k \pi$ or $\theta =3\pi/4 + 2k \pi$.

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