Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to solve the following equation: $(-12\cos^2(x)-2\cos(x)+3)(-\sin(x))=0$ on the interval $(0,2\pi)$

I found 3 out of 5 solutions which are the following: $\pi, \arccos\left(\frac{-1-\sqrt{37}}{12}\right) \mbox{ and } \arccos\left(\frac{\sqrt{37}-1}{12}\right)$

Can someone help me find the 2 other solutions

Thank you in advance

share|improve this question
    
Why do you believe there are 5 solutions? You have a trigonometric polynomial of degree 3 limited to $\,(0,2\pi)\,$, so three solutions look good for me...unless you've some further info. –  DonAntonio Oct 29 '12 at 17:41
    
I see on the graph that there exists two additional solutions –  user43418 Oct 29 '12 at 17:42
    
Oh, I see...... –  DonAntonio Oct 29 '12 at 17:46
add comment

2 Answers

up vote 2 down vote accepted

Observe that $\cos x=\cos (2\pi-x)$. So if $\alpha$ is a solution, then so is $2\pi-\alpha$.
So you have $\pi, \arccos\left(\frac{-1-\sqrt{37}}{12}\right), \arccos\left(\frac{-1+\sqrt{37}}{12}\right),2\pi-\arccos\left(\frac{-1-\sqrt{37}}{12}\right),2\pi-\arccos\left(\frac{-1+\sqrt{37}}{12}\right)$
5 solutions, as required.

share|improve this answer
    
Thank you Mr. Gulko –  user43418 Oct 29 '12 at 18:14
    
@user43418: You're welcome :-) –  Dennis Gulko Oct 29 '12 at 18:15
add comment

your answer is here,

take a look at:

wolfram alpha gives this answer

share|improve this answer
    
but copy all of the thing, sorry (I should tell it beforehand) –  kemal acikgoz Oct 29 '12 at 17:58
add comment

Your Answer

 
discard

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.