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


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You forgot to show your work...(hint: $\pi > 1$) – The Chaz 2.0 May 2 '12 at 5:01
^Orphaned by the recent edit. – The Chaz 2.0 May 2 '12 at 5:09
up vote 2 down vote accepted

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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The question has changed. Boooooooooooo! – The Chaz 2.0 May 2 '12 at 5:09
@TheChaz: Thanks for the ping. – Brian M. Scott May 2 '12 at 5:15
the answer is $\frac{\pi}{8}$ but I got $\frac{\pi}{4}$ – Sb Sangpi May 2 '12 at 5:34
@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.) – Brian M. Scott May 2 '12 at 5:37

$${}{}{}{}{}\varnothing {}{}{}{}$$ 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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The original answer must set some sort of record for concision! – Brian M. Scott May 2 '12 at 5:09
@Brian: Not anymore! (And there is a gold-badged answer of - essentially - one character...inspirational!!) – The Chaz 2.0 May 2 '12 at 5:10

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