tanx - 3/tanx = 2[3tanx-tan³x]/[1-3tan²x]

so [tan²x - 3]/tanx = 2tanx[3-tan²x]/[1-3tan²x]

so [tan²x - 3][1-3tan²x] = 2tan²x[3-tan²x]

so [tan²x - 3][1-3tan²x] + 2tan²x[tan²x-3] = 0

so [tan²x - 3][1-tan²x] = 0

so tanx = 1, -1, √3, -√3

now check the values satisfying last eq.

then put these values in 2nd eq. to get answer.