Question

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

Answer 1

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

Explanation:

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