- #1
Alettix
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- 11
Homework Statement
The following equation is to be solved for all x:
## \cos(x) + \cos(3x) = \sin(x) + \sin(3x)##
Homework Equations
The tripple angle formulas:
## \cos(3x) = 4\cos^3(x) - 3\cos(x) ##
##\sin(3x) = 3\sin(x) - 4\sin^3(x) ##
The Pythagorean trig identity:
## \sin^2(x) + \cos^2(x) = 1 ##
The Attempt at a Solution
Applying the tripple angle identities we have:
## \cos(x) + 4\cos^3(x) - 3\cos(x) = \sin(x) + 3\sin(x) - 4\sin^3(x) ##
Simplifying:
## 4\cos^3(x) - 2\cos(x) = 4\sin(x) - 4\sin^3(x) ##
## 2\cos(x)(2\cos^2(x) - 1) = 4\sin(x)(1 - \sin^2(x) ##
With the Pythagorean identity:
## \cos(x)(\cos^2(x) - \sin^2(x)) = 2\sin(x)\cos^2(x)##
Now, from this it look as if ##cos(x) = 0## should be a solution, which yields ##x_1 = \pi n## where ##n## is an integer. Continuing with the rest:
## (\cos^2(x) - \sin^2(x)) = 2\sin(x)\cos(x)##
## \cot(x) - \tan(x) = 2 ##
## \tan^2(x) + 2\tan(x) -1 =0##
Solving the second degree equation yields ##\tan(x) = -1 \pm \sqrt{2} ##, which gives ##x_2 = \frac{\pi}{8} +\pi n## and ##x_3 = \frac{-3\pi}{8} +\pi n##.
Now the only problem is that none of these solutions is right! Here https://www.desmos.com/calculator/dvbz4qpadt I plotted the functions and searched their intersection, and it doesn't match my solution. Where is my misstake? How can I solve the equation properly?
Thank you very much in advance!
