Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
I think this is very straight forward. What did you try ? – Amr Feb 8 '13 at 12:19
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
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 $n\in \mathbb{Z}$.

share|cite|improve this answer
$cos(x)=cos(y)=>x=(+ou−)y+2npi$ 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


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

share|cite|improve this answer
That'll work, but, surprisingly, may be the hard way. – Gerry Myerson Feb 8 '13 at 12:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.