Find the locations where the slope = 0

[Feodalherren]

Homework Statement

I just had a test I can't remember exactly how the question went but it was a relatively easy differentiation problem. I can only remember my result, which was

y' = 2sinxcosx - sinx

Homework Equations

The Attempt at a Solution

I noticed that 2sinxcosx is also sin2x so you could rewrite it as Sin2x - Sinx

Now, to find the places where the slope is 0 i obviously just set it to 0 and solve. My problem was that I didn't know if I should go the Sin2x - Sinx route or Factor out a sine and do it like thisL

sinx(2cosx -1) = 0

cosx = 1/2

x = ∏/3 +2∏k

x = 5∏/3 + 2∏k

where k is any integer


The above one is the route I chose to take at the end. However, if I solve for sin2x - sin x = 0

I can just say, ∏k, where k is any integer? Where is my logic flawed?

 

[Dick]

Homework Statement

I just had a test I can't remember exactly how the question went but it was a relatively easy differentiation problem. I can only remember my result, which was

y' = 2sinxcosx - sinx

Homework Equations

The Attempt at a Solution

I noticed that 2sinxcosx is also sin2x so you could rewrite it as Sin2x - Sinx

Now, to find the places where the slope is 0 i obviously just set it to 0 and solve. My problem was that I didn't know if I should go the Sin2x - Sinx route or Factor out a sine and do it like thisL

sinx(2cosx -1) = 0

cosx = 1/2

x = ∏/3 +2∏k

x = 5∏/3 + 2∏k

where k is any integer


The above one is the route I chose to take at the end. However, if I solve for sin2x - sin x = 0

I can just say, ∏k, where k is any integer? Where is my logic flawed?

x=pi*k aren't the only solutions to sin(2x)-sin(x)=0. They are just some of them. You've found the others in your first analysis. If sin(x)*(2*cos(x)-1)=0 then either sin(x)=0 or 2*cos(x)-1=0.

 

[Feodalherren]

Thanks for the answer though.