What is the general solution of the equation: `tan^2 theta + 2sqrt3 tan theta = 1?`

Answer

`theta=n pi + pi/12, n in Z or theta = mpi - (5pi)/12; m in Z`

Explanation

`tan^(2) theta + 2 sqrt3 tan theta = 1`
or `(tan theta + sqrt3)^(2) = 4`
or `tan theta = 2 - sqrt3 or -2 - sqrt3`
or `tan theta = tan 15^(@), - tan 75^(@)`
`= "tan"(pi)/(12), tan (-(5pi)/(12))`
From `tan theta = "tan" (pi)/(12)`, we get
`theta = n pi + (pi)/(12), n in Z`
From `tan theta = tan (-(5pi)/(12))`, we get
`theta = m pi - (5pi)/(12), m in Z`