SOLUTION: By substituting x=2cosy into cos3y= 4cos^3y - 3cosy , show that the equation (x)^3 - 3x - 1 = 0 has roots 2cos20°, -2sin10° and -2cos40°

So we substitute 



for the first two terms in



and get





We'll just get the answers for y in [0°,360°).
That means we get the answers for 3y in [0°,1080°)



Since x = 2cos(y),

and since cos(20°)=cos(340°), cos(140°)=cos(220°), and cos(100°)=cos(260°),

we only have 3 solutions for x,

2cos(20°), 2cos(100°), 2cos(140°)

The first solution is already in the given form.  The other two are not, so
we'll need to work on them. For the second solution, we use cos(θ) = sin(90°-θ),
and sin(-θ) = -sin(θ):

2cos(100°) = 2sin(90°-100°) = 2sin(-10°) = -2sin(10°) 

For the third solution, we use cos(θ) = -cos(180°-θ)

2cos(140°) = -2cos(180°-140°) = -2cos(40°) 

Edwin