How do you find the general solutions for #cos (x)^4-sin (x)^4=2cos (x)cos(2x)#?

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Answer 1 · 2015-10-19T23:57:21

There are 6 solutions over the interval #[0,2pi]#. The "general solution can be found by adding #2pi# to each of those 6 solutions.

I'll exclude the variable x for simplicity:

#cos^4 - sin^4=2coscos(2)#

Now simplify:

#(cos^2-sin^2)(cos^2+sin^2)=2cos(cos^2-sin^2)#

#(cos^2-sin^2)[(cos^2+sin^2)-2cos]=0#

#(cos-sin)(cos+sin)(1-2cos)=0#

Now, we have 3 factors, when set equal to 0, will give us two unique solutions each for a total of 6 solutions over the interval #[0,2pi]#

#cos-sin=0#; Solutions: #pi/4, (5pi)/4#
#cos+sin=0#; Solutions: #(3pi)/4, (7pi)/4#
#1-2cos=0#; Solutions: #(pi)/3, (5pi)/3#

Again, the general solution is simply adding/subtracting multiples of #2pi# to each of these 6 solutions.

hope that helped

How do you find the general solutions for #cos (x)^4-sin (x)^4=2*cos (x)*cos(2x)#?