The horizontal tangent line occurs at x = pi/6 and x = 5pi/6.

[Neophyte]

Homework Statement

Set f(x) = cos^2(x) + sin(x)
Find the numbers x between 0 and 2pi where the tangent line to the graph of f is horizontal.

Homework Equations

pi/6 pi/2 3pi/2 11pi/6 = x

I do not know how though

The Attempt at a Solution

-2sinx + cosx = 0
Equivalent to zero because derivative = slope = 0 because horizontal

Perhaps I have done it wrong lol, but I have no clue what to do from here if I did do it correct,

 

[Physics_Math]

Your problem is that the derivative of cos²x is not 2*sinx. remember the chain rule...

 

[Neophyte]

So it is sin^2(x) = cos(x)= 0

then -cos^2(x) +cos(x) +1 = 0

or

2cos * -sin + cosx

-2(sin)(cos) + cosx
-2sin - cos = 0

Lol uh oh ; (

 

[Mark44]

So it is sin^2(x) = cos(x)= 0

then -cos^2(x) +cos(x) +1 = 0

or

2cos * -sin + cosx

-2(sin)(cos) + cosx
-2sin - cos = 0

Lol uh oh ; (

It's very difficult to follow what you're doing, particularly when you start with an equation, and then come up with some random appearing expression whose value you don't show.

How did you get sin^2(x) = cos(x) = 0?

Start with f(x) = cos^2(x) + sin(x)
Find f'(x). (I.e., f'(x) = ...)
Set f'(x) = 0.
Solve for x.

 

[diazona]

2cos * -sin + cosx

That's the idea...

 

[Neophyte]

cos^2(x) +sin(x)

f' = -2(cos(sin) + cosx = 0

(cosx(-2(sinx) +1) = 0

cosx = 0

pi/2, 3pi2

sinx= 1/2

pi/6, 11pi/6

Is this done correctly?

 

[Cyosis]

Almost, but try to write mathematical expressions down correctly. Putting the brackets in the wrong places and forgetting arguments of functions will cause you to make mistakes and people that try to help you a headache.

For example cos(sin) means nothing, but what do you mean with it? cos(sin(x)), or cos(x)sin(x)? I know that in this case you mean cos(x)sin(x), but the meaning of something should not be guessed in mathematics.

You have f'(x)=-2 \cos x \sin x+\cos x=0.

As for your answers, the cosine part is correct. The sine part is not, pi/6 is correct, but 11/pi/6 lies in the fourth quadrant and is -1/2 instead. The other value where sin(x)=1/2 lies in the second quadrant. What is it?