I already know my first derivative to be:

$\displaystyle f'(x) = 16\cos {x}(-\sin {x} - 1)$

So I solved for my second derivative and got:

$\displaystyle -16\sin {x}(-\sin {x} - 1) + 16\cos {x}(-\cos {x})$

$\displaystyle = 16\sin {x}(\sin {x} + 1) - 16\cos^2 {x}$

I'm looking for where $\displaystyle f$ is concave up and concave down, and I figured that inflection points were a good way to find that. So, I set my equation equal to zero:

$\displaystyle 16\sin {x}(\sin {x} + 1) - 16\cos^2 {x} = 0$

$\displaystyle 16\sin {x}(\sin {x} + 1) = 16\cos^2 {x}$

$\displaystyle \sin {x}(\sin {x} + 1) = \cos^2 {x}$

I am completely stuck here. How do I solve for $\displaystyle x$ now?