Which Points on the Curve y = sin(2x) + 2 sin(x) Have a Horizontal Tangent Line?

[ahazen]

Find the x-coordinate of all points on the curve y = sin(2x) + 2 sin(x) at which the tangent line is horizontal. Consider the domain x = [0,2π).

f'(x)=2cos2x+2cosx

 

[zhermes]

Whats the slope of a horizontal line?

 

[ahazen]

the slope is zero

 

[MysticDude]

The first thing that I would want to do is to factor out the 2.

2(cos(2x) + cos(x)) this makes things easier because now all we have to focus on is the cos(2x) + cos(x) part. Why? Because the 2 in the equation is not going to make the zero, it's what ever that was inside (the cos(2x) + cos(x) in out case).

So now we just have to find what values make cos(2x) + cos(x) = 0.
The trick in this one was to find out, what x value multiplied in the first quadrant by 2 would make the (2x) part be in a quadrant with the opposite value.
We do this so the cos(x) values cancel each other out. I had to do some trial and error here and found some values. They included π/3, π, and (5π)/3.

I hope you understand my logic!

[PLAIN]http://img31.imageshack.us/img31/3142/math2r.png

 

[Mentallic]

Nice diagram mystic

If you're trying to find where $$cos(2x)+cos(x)=0$$, you can use the formula $$cos(2x)=2cos^2(x)-1$$ and then you have a quadratic in cos(x) which you can solve.

 

[MysticDude]

Nice diagram mystic

If you're trying to find where $$cos(2x)+cos(x)=0$$, you can use the formula $$cos(2x)=2cos^2(x)-1$$ and then you have a quadratic in cos(x) which you can solve.

Thanks.

Also, I never thought of making it into a quadratic. Nice trick :P

 

[Mentallic]

Yep I invented it, so don't believe anyone that tells you this trick has been known for centuries now.

 

[MysticDude]

Well when you get into limits, integrals, and derivatives the trig identities leave your brain :P

 

[ahazen]

Thank you so much everyone:) I really appreciate it:)