Solve: tan²θ-(1+√3)tanθ+√3=0?

2 Answers
Jun 22, 2018

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

Jun 22, 2018

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.