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Calculate all the values of $x$ between $0$ and $2\pi$, without using calculator. $2\sin 2x=\sqrt2$

thanks.

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    ^Orphaned by the recent edit.2012-05-02

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Okay, now you have $\sin 2x=\frac1{\sqrt2}$. You should know an angle $\theta$ such that $0\le\theta<\pi/2$ and $\sin\theta=\frac1{\sqrt2}$; if it doesn't immediately occur to you, think of the right triangles whose angles you know. There's only one other angle between $0$ and $2\pi$ whose sine is $\frac1{\sqrt2}$; what is it? (It helps here to be familiar with the circle approach to sines and cosines, but you can also get it by considering the graph of $y=\sin x$.)

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    @SbSangpi: That may be partly my fault: I wrote $\sin x$ instead of $\sin 2x$. You're right that $\sin\frac{\pi}4=\frac1{\sqrt2}$, but that means that $2x=\frac{\pi}4$, and hence $x=\frac{\pi}8$. (Now don't forget the other angle that satisfies the equation.)2012-05-02
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${}{}{}{}{}\varnothing {}{}{}{}$ Oh come on!

The solution to the revised question comes from finding angles with a since ratio of $\dfrac{\sqrt2}{2} = \dfrac{1}{\sqrt2}$. Write out two periods of such solutions (why?) and then divide the angles by two.

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    @Brian: Not anymore! (And there is a gold-badged answer of - essentially - one character...inspirational!!)2012-05-02