Double angle & half angle

[alyren]

how to solve

cos(2a) - cos(4a)=0

so far i got
cos2a - 2 cos^2(2a)-1 =0
set cos2a = y
y - 2y^2 -1 = 0
change sign
2y^2 - y + 1 =0
is this factor-able? stuck on this step!

 

[mmm4444bot]

cos2a - 2 cos^2(2a) - 1 = 0

y - 2y^2 - 1 = 0

These two lines are not correct.

On the first line, you forgot grouping symbols around the identity for cos(4a). That line should be:

cos(2a) - [2 cos(2a)^2 - 1] = 0

On the second line, subtracting -1 leads to +1, like so:

y - 2y^2 + 1 = 0

This polynomial factors nicely.

Thank you for showing your work.