Solve: tan²θ-(1+√3)tanθ+√3=0?
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"Question #f98b4"
#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#
#t = pi/4 + kpi#
#t = pi/3 + kpi#
#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.
