1. ## Inflection Points

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?

2. Assuming everything you've done so far is correct:
$\displaystyle sinx(sinx+1) = cos^{2}x$

$\displaystyle sin^{2}x + sinx = 1 - sin^{2}x$

$\displaystyle 2sin^{2}x + sinx - 1 = 0$

3. The sign of the second derivative will indicate which direction the function opens: positive second derivative means concave up, negative second derivative means concave down. So those intervals on which the function has a positive second derivative are the intervals in which the function is concave up, vice versa for points where the function has a negative second derivative.

4. Originally Posted by o_O
Assuming everything you've done so far is correct:
$\displaystyle sinx(sinx+1) = cos^{2}x$

$\displaystyle sin^{2}x + sinx = 1 - sin^{2}x$

$\displaystyle 2sin^{2}x + sinx - 1 = 0$

How do you solve the quadratic when it involves trig?

5. Treat it like a polynomial in $\displaystyle \sin(x)$. If you like it better say $\displaystyle \sin(x) = y$.

Then your quadratic becomes $\displaystyle 2y^2+y-1 = 0$

once you solve for the possible values of $\displaystyle y$ substitue $\displaystyle \sin(x)$ back in and solve for $\displaystyle x$ using $\displaystyle \arcsin$ or some other technique.

6. Originally Posted by iknowone
Treat it like a polynomial in $\displaystyle \sin(x)$. If you like it better say $\displaystyle \sin(x) = y$.

Then your quadratic becomes $\displaystyle 2y^2+y-1 = 0$

once you solve for the possible values of $\displaystyle y$ substitue $\displaystyle \sin(x)$ back in and solve for $\displaystyle x$ using $\displaystyle \arcsin$ or some other technique.
So I would have:

$\displaystyle (2y-1)(y+1)=0$

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

So x is 0 when $\displaystyle \sin {x} = -1, \frac {1}{2}$

Thanks guys I appreciate it.