<s>[INST] Finding minima and maxima of $\sin^2x \cos^2x$ I am trying to find the turning points of $\sin^2x \cos^2x$ in the range $0<x<\pi$.
So far I have:
$\begin{align} f'(x) & =\cos^2x \cdot 2\sin x \cdot \cos x + \sin^2x \cdot 2\cos x \cdot -\sin x \\
& = 2 \cos^3x \sin x - 2\sin^3x \cos x \\
& = 2 \sin x \cos x (\cos^2x - \sin^2x) \\
& = 2 \sin x \cos x (\cos x + \sin x) (\cos x - \sin x) 
\end{align}$
And, after lots of simplification, I think I've found that:
$f''(x) = 2 \left[\left( \cos^2x-\sin^2x \right)^2 - 4\sin^2x \cos^2 x\right]$
My questions are:


*

*How can I evaluate $0 = \cos x + \sin x$ and $0 = \cos x - \sin x$ without resorting to graph plotting?

*Are there trigonometric identities that I could have used to simplify either derivative?

 [/INST] Hint:
$$\begin{align}
0&=\cos x+\sin x\\
\sin x &= -\cos x \\
\text{Divide}&\text{ both sides by }\cos x\\
\tan x &= -1
\end{align}$$
Now you can solve for $x$ using inverse trigonometric functions.
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