# Is there is quicker way to find the solutions to cos(x) = - cos(3x)?

Asked by PhiNotPi (12647) March 27th, 2014

This was a math problem I faced in class earlier today, but I felt that my method of solving it was definitely the long way:

cos(x) = -cos(3x)
cos(x) + cos(3x) = 0
cos(x) + cos(x+2x) = 0
cos(x) + cos(x)*cos(2x) – sin(x)*sin(2x) = 0
cos(x) + cos(x) * (cos(x)^2 – sin(x)^2) – sin(x) * (2*sin(x)*cos(x)) = 0
cos(x) + cos(x)^3 – cos(x)*sin(x)^2 – 2*sin(x)^2*cos(x) = 0
cos(x) * (1 + cos(x)^2 – sin(x)^2 – 2*sin(x)^2) = 0
cos(x) * (2*cos(x)^2 – 2*sin(x)^2) = 0
cos(x) * 2 * cos(2x) = 0

From here, we can find the solutions, which are pi/4, 2pi/4, 3pi/4, 5pi/4, 6pi/4, and 7pi/4.

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That’s all fine, but it’s simply too long. I feel that there just has to be an easier way. Simpler is better. After all, I’m the sort of guy who likes the origami one-fold stegosaurus.

Are there any helpful trigonometry tricks which would shorten the amount of work I have to do?

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3x = pi – x + 2npi or
3x = pi + x + 2npi

x = (2n+1)pi/4 or
x=(2n+1)pi/2

@LostInParadise‘s solution is of course the fastest I can think of. You could also bring both terms to one side and use formulas for turning addition to multiplication and go from there.

cos x + cos y =2 cos ((x+y)/2) cos((x-y)/2)

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