
Andy C. answered 08/19/18
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Indeed, cos3x = cos(2x+x) = cos(2x)cosx - sin2x sinx <--- angle addition for cosine
= (cos^2 - sin^2)cos - (2 sin cos) sin <---- trig identities for cos2x and sin(2x);
trig arguments are now just x, so it is dropped
for the sake of convenience
= cos^3 - sin^2*cos - 2 sin^2 cos
= cos^3 - 3 sin^2 cos
= cos^3 - 3 ( 1 - cos^2) cos <---- sin^2 + cos^2 = 1 so sin^2 = 1 - cos^2
= cos^3 - 3 (cos - cos^3)
= cos^3 - 3 cos + 3 cos^3
= 4cos^3 - 3cos
which proves the identity shown in the second line
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Now to solve the equation in the first line....
1 + 4 cos^3 - 3cos = cos + cos^2 <--- substitutes the identity for cos 3x
1 + 4 cos^3 - 3cos - cos - cos^2 = 0 <---- everything to left side
4 cos^3 - cos^2 - 4cos + 1 = 0 <--- Combines Like Terms CLT
4cos^3 - cos^2 - (4cos - 1) = 0 <--- factors out -1 from last two terms
cos^2 (4cos - 1) - (4cos-1) = 0 <---- factors out cos^2 from first two terms
(cos^2 -1)(4 cos-1) = 0 <---- factors out 4cos-1 : factors by grouping
(cos+1)(cos-1)(4cos-1) = 0 <---- factors difference of squares pattern N^2-1 = (N+1)(N-1)
By zeros product property:
cos = 1/4 OR cos = +or- 1
The cosine is 1/4 when inverse-cosine(0.25) = 75.52248781407....
cosine is positive in quadrants 1 and 3
The angles are 75.52248781407.... and 180+75.52248781407 = 255.52248781407...
cosine is 1 at 0 degrees and -1 at 180 degrees
So the solution set is { 75.52248781407...., , 255.52248781407..., 0, 180}
and of course, these angles are in degrees.