**Q: Let f(x)= 3sin ^{2}x + 4cos^{3}x. Determine where the tangent line is horizontal.**

A: To have a horizontal (flat) tangent line, it means your slope is 0. Therefore, we must take the derivative (slope) and set it equal to 0.

f(x)= 3sin^{2}x + 4cos^{3}x

When taking the derivative, we must invoke the chain rule, which I like to describe as: the derivative of the outside, times the derivative of the inside.

So, the derivative of (something)^{2} = 2(something unchanged) * (derivative of that something)

f'(x)= 3sin^{2}x + 4cos^{3}x = 2(3sinx)*cosx + 3(4cos^{2}x)*(-sinx)

f'(x) = 6sinx(cosx) – 12cos^{2}x(sinx)

Now, we want to set the derivative = o

0 = 6sinx(cosx) – 12cos^{2}x(sinx)

[factor the right side like so]:

0 = 6sinx(cosx) [1 - 2cosx]

So, either (1) 0 = 6sinx(cosx) **OR **(2)** **0 = [1 - 2cosx]

**(1)** 0 = 6sinx(cosx) means that either sinx = 0 or cosx = 0

sinx = 0 when **x = 2Πk , where k is any integer
**

cosx = 0 when **x = ****Π/2 + ****Πk , where k in any integer**

**(2) **0 = [1 - 2cosx]

2cosx = 1

cosx = 1/2

This happens when **x = ** **Π/3 + 2****Πk OR 5****Π/3 + 2****Πk**

So, putting all of our answers together… We know that the tangent line is horizontal when:

**x =** **2Πk, ****Π/2 + ****Πk, ****Π/3 + 2****Πk OR 5****Π/3 + 2****Πk**

**Wow! Long problem? Confused? Let me know.
**