1 + cos x + cos 2x = 0

1 + cos(x) + cos(2x) = 0
1 + cos(x) + cos(x + x) = 0
1 + cos(x) + cos(x)^2 - sin(x)^2 = 0
1 + cos(x) + cos(x)^2 - (1 - cos(x)^2)) = 0
1 + cos(x) + 2cos(x)^2 - 1 = 0
cos(x) + 2cos(x)^2 = 0
cos(x)(1 + 2cos(x)) = 0
=> cos(x) = 0, x = π/2, 3π/2
=> 1 + 2cos(x) = 0, cos(x) = -1/2, x = 2π/3, 4π/3

So x = π/2, 2π/3, 4π/3, 3π/2.

1 + cos x + cos 2x = 0
OR (1 + cos 2x) + cos x = 0
OR 2 cos^2 x + cos x = 0
OR cos x (2 cos x + 1) = 0
giving cos x = 0 = cos (pi/2) and cos x = -- 1/2 = cos (2 pi/3)
So x = 2 n pi +/-- pi/2 and x = 2 n pi +/-- 2 pi/3 ANSWER

cos=1/x+2x