Question

Solve: tan²θ-(1+√3)tanθ+√3=0?

Answer 1

`tan²θ-(1+√3)tanθ+√3=0`

`=>tan²θ-tantheta-√3tanθ+√3=0`

`=>tantheta(tanθ-1)-√3(tanθ-1)=0`

`=>(tanθ-1)(tantheta-√3)=0`

So `tantheta=1=tan(pi/4)`

`=>theta=npi+pi/4" where "n inZZ`

Again

`tantheta=sqrt3=tan(pi/3)`

`=>theta=kpi+pi/3 " where " k in ZZ`

Answer 2

`t = pi/4 + kpi`
`t = pi/3 + kpi`

Explanation:

`f(t) = tan^2 t - (1 + sqrt3)tan t + sqrt3 = 0`.
Solve this quadratic equation for tan t.
Since (a + b + c = 0), use shortcut. The 2 real roots are:
tan t = 1 and `tan t = c/a = sqrt3`
a. `tan t = 1`
Trig table and unit circle give --> `t = pi/4 + kpi`
b. `tan t = sqrt3`
Trig table and unit circle give --> `t = pi/3 + kpi`.
Check.
`t = pi/3` --> `tan^2 t = 3` --> `- (1 + sqrt3)tan t = - sqrt3 - 3`.
`f(pi/3) = 3 - sqrt3 - 3 + sqrt3 = 0`. Proved.
`t = pi/4` --> `tan^2 t = 1` --> `- (1 + sqrt3)tan t = - 1 - sqrt3`
`f(pi/4) = 1 - 1 - sqrt3 + sqrt3 = 0`. Proved.