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 am having a problem with the following function:


I need to examine the sign of $f'(x)$

I noticed that f is $2\pi$-periodic, therefore we need to analyze f(x) on $[0;2\pi]$

In addition to that $f(x)=-4\cos^3(x)-\cos^2(x)+3\cos(x)+1$

Hence $\forall x \in ]0;2\pi[, f'(x)=(-12\cos^2(x)-2\cos(x)+3)(-\sin(x))$

Let us solve $f'(x)=0$

We have $\forall x \in ]0;2\pi[ -\sin(x)=0 \Leftrightarrow x=\pi$

For the second equation, I know that there are two solution on $]0;2\pi[$ (since I visualized it on the graph), but I am unable to determine them by calculuation.

Please help.

Thank you in advance

share|improve this question

2 Answers 2

up vote 1 down vote accepted

You forgot to decrement the power of the $\cos^3x$ term. Once you fix that, you should have a quadratic you can solve for $\cos x$.

share|improve this answer
Ah ! Thank you!!! –  user43418 Oct 29 '12 at 15:04
I have a question. I determined the 3 out of 5 zeros of $]0;2\pi[$ which are: $\pi$; $Arccos(\frac{-1-\sqrt{37}}{12})$ and $Arccos(\frac{\sqrt{37}-1 }{12})$ However there still exists 2 more which I am unable to determine. In addition to that I know that f' is $\pi$-periodic –  user43418 Oct 29 '12 at 16:26
Can someone help me in order to determine the two additional values –  user43418 Oct 29 '12 at 16:55
@user43418 $\cos(2\pi-x)=\cos(-x)=\cos x$. Is this what you're looking for? –  Mike Oct 29 '12 at 18:05

$$ \begin{align} f'(x)&=2 \cos x\sin x +3 \sin (3x) = 2 \cos x \sin x +3 (3 \sin x - 4 \sin^3 x) \\ &= \sin x \left( 2 \cos x + 9 -12 \sin^2 x \right) = \sin x \left( 2 \cos x + 9 -12 (1-\cos^2 x) \right) \end{align} $$ so that you have $\sin x=0$ and $12 \cos^2 +2 \cos x -3=0$.

share|improve this answer

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.