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Find all solutions in the interval [0,2pi): tan^2x+(sqrt3 - 1)tanx-sqrt3 = 0

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1 Answer

f(x) = tan2(x) + (√3 - 1)[tan(x)] - √3 = 0
 
tan2(x) + √3[tan(x)] - tan(x) - √3 = 0
 
Factor into
 
[-1 + tan(x)]*[√3 + tan(x)] = 0
 
which means
 
[-1 + tan(x)] = 0 and/or [√3 + tan(x)] = 0
 
Then
 
tan(x) = 1
 
tan-1(1) = pi/4 radians
 
For the other equation
 
[√3 + tan(x)] = 0
 
tan(x) = -√3
 
tan-1(-√3) = -pi/3
 
so that
 
x = pi/4 or -pi/3 in the interval [0, 2pi]