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# Advanced Math/position of the nodes on a waveform

Question
I have a waveform of y1(x)= 1.2*cos(1.2*x)+1.8*cos(1.8*x), and i need to find the position of the nodes.

I found the nodes for y2(x)= sin(x*1.2)+sin(x*1.8) with either x = nπ/(1.5) or x = (2n+1)*(π/0.6), where n = 0, 1,2 3, and so on.

but i can't find an easy way for the first one.

If we let t = 0.6x, then x = t /0.6 and we get y1(t/0.6) = 1.2*cos(2t) + 1.8*cos(3t).

Knowing that cos(a+b) = cos(a)cos(b) - sin(a)sin(b) with a=t and b=2t, we get
so cos(3t) = cos(2t)cos(t) - sin(2t)sin(t), which giveds
y1(t/0.6) = 1.2*cos(2t) + 1.8*cos(t)*cos(2t) - 1.8*sin(t)sin(2t).

Factoring out the cos(2t) from the first two terms gives
y1(5t/3) = (1.2 + 1.8*cos(t))cos(2t) - 1.8*sin(t)sin(2t).

It is also known that cos(2t) = 2*cos²(t) - 1 and sin(2t) = 2*sin(t)cos(t).
This gives y1(5t/3) = (1.2 + 1.8*cos(t))(2*cos²(t) - 1) - 1.8*sin(t)*2*sin(t)cos(t).

The 1st term of y1 is (1.2 + 1.8*cos(t))(2*cos²(t) - 1), and this is
3.6*cos³(t) + 2.4*cos²(t) - 1.8*cos(t) - 1.2.

That 2nd term is 1.8*sin(t)*2*sin(t)cos(t), and this is
3.6*sin²(t)cos(t) = 3.6*(1-cos²(t))cos(t) = 3.6*cos(t)-3.6*cos³(t)).

Combining the 1st and 2nd terms cancels the 3.6*cos³(t), giving
2.4*cos²(t) + 1.8*cos(t) - 1.2.

Letting u = cos(t), this is a quadratic in u, which is 2.4u² + 1.8u - 1.2.
Rewriting this gives (3/5)(4u² + 3u - 2).

The solution to 4u² + 3u - 2 is given by u = (-3±5)/8, which gives u=1/4 or u=-1.

Knowing that u=cos(t), this can be used to find t.
Note that if u=-1, t=(2n+1)π for any integer n and also
if u=1/4, t = arccos(1/4), which gives another set of numbers that are 2π apart.

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#### Scott A Wilson

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Master of Science at OSU with high honors in mathematics. Bachelor of Science at OSU with high honors in mathematical sciences. This degree involved mathematics, statistics, and computer science. I also took sophmore level physics and chemistry while I was attending college. On the side I took raquetball, but that's still not relevant.

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