# What are the points of inflection, if any, of f(x) = 5cos^2x − 10sinx  on x in [0,2pi]?

Sep 16, 2016

$x = \frac{\pi}{6} , \frac{5 \pi}{6} , \frac{3 \pi}{2}$

#### Explanation:

We have: $f \left(x\right) = 5 {\cos}^{2} \left(x\right) - 10 \sin \left(x\right)$; $x \in \left[0 , 2 \pi\right]$

In order to determine the points of inflection, we need to evaluate the second derivative of the function:

$\implies f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(5 {\cos}^{2} \left(x\right)\right) - \frac{d}{\mathrm{dx}} \left(10 \sin \left(x\right)\right)$

$\implies f ' \left(x\right) = \left(5 \cdot - \sin \left(x\right) \cdot 2 \cos \left(x\right)\right) - \left(10 \cdot \cos \left(x\right)\right)$

$\implies f ' \left(x\right) = - 10 \sin \left(x\right) \cos \left(x\right) - 10 \cos \left(x\right)$

$\implies f ' ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(- 10 \sin \left(x\right) \cos \left(x\right) - 10 \cos \left(x\right)\right)$

$\implies f ' ' \left(x\right) = \left(- 10 \cdot \left(\sin \left(x\right) \cdot - \sin \left(x\right) + \cos \left(x\right) \cdot \cos \left(x\right)\right)\right) - \left(10 \cdot - \sin \left(x\right)\right)$

$\implies f ' ' \left(x\right) = - 10 \left(- {\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)\right) + 10 \sin \left(x\right)$

$\implies f ' ' \left(x\right) = - 10 \left({\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)\right) + 10 \sin \left(x\right)$

The points of inflection occur at those points where the second derivative is equal to zero:

$\implies f ' ' \left(x\right) = 0$

$\implies - 10 \left({\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)\right) + 10 \sin \left(x\right) = 0$

One of the Pythagorean identities is ${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$.

We can rearrange this to get:

$\implies {\cos}^{2} \left(x\right) = 1 - {\sin}^{2} \left(x\right)$

Let's apply this rearranged identity to get:

$\implies - 10 \left(\left(1 - {\sin}^{2} \left(x\right)\right) - {\sin}^{2} \left(x\right)\right) + 10 \sin \left(x\right) = 0$

$\implies - 10 \left(1 - 2 {\sin}^{2} \left(x\right)\right) + 10 \sin \left(x\right) = 0$

$\implies - 10 + 20 {\sin}^{2} \left(x\right) + 10 \sin \left(x\right) = 0$

Let's rearrange this to form a quadratic equation:

$\implies 20 {\sin}^{2} \left(x\right) + 10 \sin \left(x\right) - 10 = 0$

$\implies 10 \left(2 {\sin}^{2} \left(x\right) + \sin \left(x\right) - 1\right) = 0$

$\implies 2 {\sin}^{2} \left(x\right) + \sin \left(x\right) - 1 = 0$

Then, let's factorise using the middle-term break:

$\implies 2 {\sin}^{2} \left(x\right) + 2 \sin \left(x\right) - \sin \left(x\right) - 1 = 0$

$\implies 2 \sin \left(x\right) \left(\sin \left(x\right) + 1\right) - 1 \left(\sin \left(x\right) + 1\right) = 0$

$\implies \left(\sin \left(x\right) + 1\right) \left(2 \sin \left(x\right) - 1\right) = 0$

We now have a product that is equal to zero, so either one of the multiples must be equal to zero:

$\implies \sin \left(x\right) + 1 = 0$

$\implies \sin \left(x\right) = - 1$

$\implies x = \arcsin \left(- 1\right)$

$\implies x = \frac{3 \pi}{2} + 2 \pi n$

or

$\implies 2 \sin \left(x\right) - 1 = 0$

$\implies 2 \sin \left(x\right) = 1$

$\implies \sin \left(x\right) = \frac{1}{2}$

$\implies x = \arcsin \left(\frac{1}{2}\right)$

$\implies x = \frac{\pi}{6} + 2 \pi n , \frac{5 \pi}{6} + 2 \pi n$

However, the domain is given as $x \in \left[0 , 2 \pi\right]$.

Therefore, the points of inflection occur at $x = \frac{\pi}{6}$, $x = \frac{5 \pi}{6}$ and $x = \frac{3 \pi}{2}$.