<s>[INST] Extrema and inflection points of the function $y = \cos^2(x) - \cos(x)$ Please help me find the extrema and inflection points of the function $y = \cos^2(x) - \cos(x)$.
So far: $$y'=-2 \cos(x)\sin(x)+\sin(x)$$ $$y'' = 2\sin^2(x) - 2\cos^2(x) + \cos(x)$$
$y' = 0$ when $x = 0, \pi, \pi/3, 5\pi/3$
I plugged these values into $y''$ to find min/max which are as follows:
$x = 0, \pi$ : MAX
$x =  \pi/3$, $5\pi/3$ : MIN
The issue I have is finding the inflection points, i used the quadratic formula to solve for $\cos(x)$ by converting (via trig ID's) $y'' = 2\sin^2(x) + 2\cos^2(x) + \cos(x)$ to $y'' = -2\cos^2(x) + \cos(x) = 0$ which gave me $\cos(x) = 0, 1/2$, and $x = \pi/2, 3\pi/2, \pi/3, 5\pi/3$
The solution shows that the inflection points are at $x = 32$ deg $32', 126$ deg $23', 233$ deg $37'$, and $327$ deg $28' $
Can you please help by showing how these inflection points were solved for? Where did I make my mistake? I have no idea how they got those degrees. 
 [/INST] You changed from addition to subtraction of the first two terms in $y''$: 
Recall that $$y'' = 2sin^2(x) - 2cos^2(x) + cos(x)$$
$$y'' = 2(1-\cos^2 x) - 2 \cos^2(x) + \cos x \\
= -4\cos^2 x + \cos x + 2 \iff 4\cos^2 x - \cos x - 2$$
Try solving for $\cos x$ now, since we have a quadratic in $\cos x$. I suggest using the quadratic formula: $\cos x = \frac{1 \pm \sqrt{33}}{8}$. You can get an approximation by using $\cos^{-1}(\cos(x))$ for each of the two solutions for $\cos x$.
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