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I am having a problem with the following function:

$f(x)=\sin^2(x)-\cos(3x)$

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

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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 $\cos x$.

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

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