Trigonometric Equation with Unequal Coefficients: How to Solve?

[John O' Meara]

Solve 3\cos\frac{3x}{2}=\cos\frac{x}{2} I am interested to know how to go about solving such an equation where the coefficient of cos(x) is not equal to 1. I know how to solve \cos\theta = \cos\alpha, \mbox{ it is just} \theta = 2n\pi +/-\alpha I teach various maths subject to myself, then I realized I didn't know how to solve the above equation. Thanks.

 

[l'Hôpital]

Well, \cos \frac{3x}{2} = \cos (x + \frac{x}{2})

Therefore, you get x = (2n + 1)\pi \ for \ n = ...-3, -2, -1,0,1,2,3,... as a freebie.

 

[Gerenuk]

Surely the non-trivial solutions are more interesting.

You can always transform multiple angles to powers of trigonometric functions (and vice versa).
For example with de Moivre
\cos ny=\Re\left(\cos y+i\sin y\right)^n
\cos 3y=\cos^3y-3\cos y\sin^2y=\cos^3y-3\cosy(1-\cos^2y)=\cos y(4\cos^2y-3)=\cos y(2\cos 2y-1)

Since here on angle is three times the other, you can use y=x/2 and get
3\cos 3y=\cos y
3\cos y(2\cos(2y)-1)=\cos y
and therefore either
x=\pm\frac{\pi}{4}+k\pi
or
x=\pm\frac14\arccos\frac23+k\frac{\pi}{2}

 

[John O' Meara]

Thanks for the time you spent on working that out. I didn't know of De Moivre's. I worked it out afterwards and get 2y=\pm \arccos(\frac{2}{3})+ 2n\pi. And y=x/2, I couldn't x=\pm\frac{\pi}{4}+k\frac{\pi}{2}, which is what you got. According to my graphing calculator the graph crosses the x-axis at \arccos(\frac{2}{3}) which looks close to pi/4 but is not. It crosses the x-axis again close to pi.

 

[Gerenuk]

Not sure what you mean.

Both 1/4arccos 2/3 and pi/4 are solutions.

 

[John O' Meara]

I get a principle angle of arccos(2/3) = 2y and -arccos(2/3) is another one , right. Maybe I have already gone wrong. If I saw the steps you used I could follow you. <br /> 3\cosy(2\cos(2y)-1)=cosy, \mbox{therefore}\\<br /> 3(2\cos(2y)-1)=1, \\<br /> 2\cos(2y)-1=\frac{1}{3}, \\<br /> 2\cos(2y)=4/3, \\<br /> \cos(2y)=\frac{2}{3}

 

[Gerenuk]

The equation is
3\cos y(2\cos(2y)-1)=\cos y
Now either cos(y) is zero (which gives you pi/4 for x) or you can divide by cos(y) and continue to get x=arccos(2/3)/4

 

[John O' Meara]

Thanks.

 

[willem2]

Since here on angle is three times the other, you can use y=x/2 and get
3\cos 3y=\cos y
3\cos y(2\cos(2y)-1)=\cos y
and therefore either
x=\pm\frac{\pi}{4}+k\pi
or
x=\pm\frac14\arccos\frac23+k\frac{\pi}{2}

If y = x/2 then x = 2y, so if the solutions for y are

y = \frac {\pi} {2} + k \pi

or

y = \pm \frac {1} {2} \arccos \frac {2}{3} + k \pi

then the solutions for x will be

x = (2k + 1) \pi

or

x = \pm\arccos \frac {2}{3} + 2 k \pi

 

[John O' Meara]

Yes, that is what I finally finished up with, willem2.Thanks willem2.