How do I find the inflection points of this equation?

Question

I have an equation $f(x) = \cos^2{x} - 2\sin{x}$ and I want to find out the inflection points (where the concavity changes) and where the graph is increasing and decreasing. I'm having trouble with the inflection points.

So the first derivative is:

$$f'(x) = -2\sin{x}\cos{x} - 2\cos{x} = -2\cos{x}(\sin{x} + 1)$$

So the critical values here are when $f'(x) = 0$ so $x = {\pi/2, 3\pi/2}$. From here I can use the critical values to divide up $f'(x)$ into intervals and determine which intervals are positive and negative. Turns out that $f'$ is positive only between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ right?

So the local min is $(\frac{\pi}{2}, -2)$ and the max is $(\frac{3\pi}{2}, 2)$ right?

b) Inflection points:

$$f''(x) = -2\cos{x}(\cos{x}) + (\sin{x} + 1)(2\sin{x})$$

$$ -2\cos^2{x} + 2\sin^2{x} + 2\sin{x} $$

$$= 2(\sin^2{x} + \sin{x} - \cos^2{x})$$

How do I simplify further so I can find the zeroes? I need to find where the $f''(x) = 0$. How do I do this?

Answer 1

$\begin{array}\\ f''(x) &= 2(\sin^2{x} + \sin{x} - \cos^2{x})\\ &= 2(\sin^2{x} + \sin{x} - (1-\sin^2{x}))\\ &= 2(2\sin^2{x} + \sin{x} - 1)\\ \end{array} $

Letting $\sin(x) = z$, $f''(x) =2(2z^2+z-1) $ so $f''(x) = 0$ when $2z^2+z-1 = 0$ or $z =\dfrac{-1\pm\sqrt{1+8}}{4} =\dfrac{-1\pm 3}{4} =-1, \frac12 $.

Therefore $x = \arcsin(-1, \frac12) =3\pi/2, \pi/6 $.

Answer 2

Observe that

$$f''(x)=2(\sin^2x-\cos^2x+\sin x)=2\left(2\sin^2x+\sin x-1\right)=0\iff\ldots$$

and you have an easy quadratic to solve (if you want, substitutte $\;t:=\sin x\;$ above)