If the equation is tan(2x) - 2 cos(x) = 0:
We have
tan(2x) = sin(2x) / cos(2x)
sin(2x) = 2 sin(x) cos(x)
So, rewriting the equation gives
tan(2x) - 2 cos(x) = 0
2 sin(x) cos(x) / cos(2x) - 2 cos(x) cos(2x) / cos(2x)= 0
2 cos(x) • (sin(x) - cos(2x)) / cos(2x) = 0
The left side is 0 whenever the numerator is 0:
2 cos(x) = 0 or sin(x) - cos(2x) = 0
cos(x) = 0 or sin(x) - (1 - 2 sin²(x)) = 0
cos(x) = 0 or 2 sin²(x) + sin(x) - 1 = 0
cos(x) = 0 or (2 sin(x) - 1) (sin(x) + 1) = 0
cos(x) = 0 or 2 sin(x) - 1 = 0 or sin(x) + 1 = 0
cos(x) = 0 or sin(x) = 1/2 or sin(x) = -1
Solving each case gives 3 families of solutions:
[x = π/2 + 2nπ or x = -π/2 + 2nπ] or
[x = π/6 + 2nπ or x = 5π/6 + 2nπ] or
[x = -π/2 + 2nπ]
where n is any integer. The last family is already accounted for in the first, so
x = π/2 + 2nπ or x = -π/2 + 2nπ or x = π/6 + 2nπ or x = 5π/6 + 2nπ
