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