# Calculus Problem (involes critical numbers)

1. Mar 15, 2004

### Cod

The directions state to find any critical numbers of the function within the interval 0 < x < 2pi. The function is:

$$h(x) = sin^2(x) + cos(x)$$

I've already found that the derivative is:

$$2sin(x)cos(x) - sin(x)$$

I've set that equal to zero, and that's where I'm stuck. I guess this is more of an algebra question than anything. The answer I keep getting is:

$$tan(x)/sin(x) = 2$$

However, I don't think its correct. So any help would be greatly appreciated.

2. Mar 15, 2004

$$2\sin{x}\cos{x} - \sin{x} = 0$$
$$2\sin{x}\cos{x}= \sin{x}$$
$$2\cos{x}=0$$
$$\cos{x}=0$$

...?

3. Mar 15, 2004

### Cod

Ohhhhh, I'm an idiot. I was adding $$sin x$$ to both sides then dividing by $$cos x$$ to get $$tan x$$. Well, I guess I was just looking over the obvious the whole time.

Thanks for the assistance.

4. Mar 15, 2004

### HallsofIvy

Staff Emeritus
cod: Don't call yourself an idiot. Let's reserve that for cookiemonster! (Hey, it's not often I can do that!!)

NO, 2sin x cos x= cos x does NOT immediately lead to
"2cos x= 0"- it leads to 2cos x= 1 !

In fact, the best way to do this is to factor the original form:
2sin x cos x- sin x= sin x(2 cos x- 1)= 0 so

either sin x= 0 or 2 cos x-1= 0. That is, either sin x= 0
or cos x= 1/2. You can get x itself from those.

5. Feb 10, 2011

### pinakeenya

the answer for: Find any critical numbers of the function. (Enter your answers as a comma-separated list.)
(sin(x))2 + cos(x) 0 < x < 2π

is: Pi/3 , Pi, 5Pi/3 so you know