1. ## Horizontal tangent points

Hey, I'm back with another homework hand-in problem that I can't figure out how to do and was hoping someone could help me out here. I need to find all the points of horizontal tangent lines of

y = x + Cos(2x) where 0<= x <= 2pi

Now I was going to use 0 and 2pi as the end points, find the derivative of the problem, and then solve for x and set it = to 0. But... the derivative is

1 + -2Sin(2x) and it kills off the only x I could work with...

2. ## Re: Horizontal tangent points

Originally Posted by APMAPM
Hey, I'm back with another homework hand-in problem that I can't figure out how to do and was hoping someone could help me out here. I need to find all the points of horizontal tangent lines of

y = x + Cos(2x) where 0<= x <= 2pi

Now I was going to use 0 and 2pi as the end points, find the derivative of the problem, and then solve for x and set it = to 0. But... the derivative is

1 + -2Sin(2x) and it kills off the only x I could work with...
Why would that "kill off x"? 1- 2sin(2x)= 0 gives sin(2x)= 1/2. So 2x= what? There are two solutions between 0 and $2\pi$.

3. ## Re: Horizontal tangent points

Yes, $y^{\prime} = 1 - 2\sin(2x)$, but it certainly did not "kill off" the x. The x is still part of the angle! Set $y^{\prime} = 0$ and then you get $\sin(2x) = \frac{1}{2}$

Can you take it from there?

4. ## Re: Horizontal tangent points

Ah, I see. Thank you both for answering. I actually never did take trig so if there's some sort of property you can do with that, I wouldn't know. Can you take the sin of both sides?

5. ## Re: Horizontal tangent points

Originally Posted by APMAPM
Ah, I see. Thank you both for answering. I actually never did take trig so if there's some sort of property you can do with that, I wouldn't know. Can you take the sin of both sides?
actually, you would use the inverse sine function ... you need to research the precalculus/trigonometry topic of solving trig equations and any other aspect of trigonometry you are unfamiliar with ... this knowledge is necessary for understanding problems such as this.