# Horizontal tangent line

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

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Whats the slope of a horizontal line?

the slope is zero

MysticDude
Gold Member
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 [Broken]

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Mentallic
Homework Helper
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
Gold Member
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
Homework Helper
Yep :tongue: I invented it, so don't believe anyone that tells you this trick has been known for centuries now.

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

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