Calculus Problem (involes critical numbers)

[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.

 

[cookiemonster]

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

...?

cookiemonster

 

[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.

 

[HallsofIvy]

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.

 

[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