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How to find the values of $x$ such that $$\sqrt{\frac{1+\cos 4x}{2}}=\frac{1}{\sqrt{2}}$$

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I think this is very straight forward. What did you try ? –  Amr Feb 8 '13 at 12:19
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Square roots are a pain in the neck, aren't they? Do you know a good way to get rid of them? –  Gerry Myerson Feb 8 '13 at 12:20
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2 Answers 2

up vote 2 down vote accepted

Sqaure both sides: $$\frac{1+\cos(4x)}{2}=\frac{1}{2}$$ Multiply by two $$1+\cos(4x)=1$$ subtract one from both sides: $$\cos(4x)=0$$ $\cos(y)=0$ when $y=\frac{\pi}{2}+ n\pi$ for $n\in \mathbb{Z}$. So we have to satisfy $4x=\frac{\pi}{2}+ n\pi$ for $n\in \mathbb{Z}$. So your solution for $x$ should be $x=\frac{\pi}{8}+\frac{n\pi}{4}$ where $n\in \mathbb{Z}$.

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$cos(x)=cos(y)=>x=(+ou−)y+2npi$ you forgot $2$ .And how about the second case ,the minus (-)? –  Frank Feb 8 '13 at 21:54
    
Since $n \in \mathbb{Z}$ it can be both negative or non-negative. Furthermore the question is not concerned with $\cos(x)=\cos(y)$, I merely temporarily introduced the $y$ to make it a little clearer that for this question $y=4x=\frac{\pi}{2}+ n\pi$. –  Slugger Feb 9 '13 at 2:33
    
yes but it's not correct in other situation. –  Frank Feb 9 '13 at 11:46
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Hint:

$$\frac{1+\cos(2x)}{2} =\cos^2 x$$

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That'll work, but, surprisingly, may be the hard way. –  Gerry Myerson Feb 8 '13 at 12:22
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