Question

Hey guys! Does anybody know the answer to tan2x-2cosx=0

Answer 1

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