(Q&A) Solve: cos2x+cos4x=0 ?

Q:-

Solve between (0, 2pi): cos2x +cos4x = 0

A:-

0 = cos2x + cos4x = cos2x + 2(cos2x)^2 - 1
because cos2a = cos^2 a - sin^2 a = cos^2 a - (1 - cos^2 a)

Put A=cos2x.
Then from the above:
0 = 2A^2 + A - 1 = (2A - 1)(A + 1)
and this implies
A = -1 or A = 1/2
ie
cos2x = -1 or cos2x = 1/2

cos2x = -1 implies:
2x = pi or 3pi, so x=pi/2 or 3pi/2

cos2x = 1/2 implies:
2x = pi/3 or 5pi/3 or 7pi/3 or 11pi/3
so x=pi/6 or 5pi/6 or 7pi/6 or 11pi/6

So all the solutions are:
pi/6, pi/2, 5pi/6, 7pi/6, 3pi/2, 11pi/6