0
sin(60) or sin(PI/3) = sqrt(3)/2
cos(60) or cos(PI/3)=1/2
tan(60) or tan(PI/3) = sin(60)/cos(60)=sqrt(3)
But we want tan for -sqrt(3). Tangent is negative in quadrant II and IV.
In Quadrant IV, we compute 360-60=300 or 2PI-PI/3 =5PI/3
tan(5PI/3) = -sqrt(3)
Tangent is also negative in the second quadrant, so we compute PI-PI/3=2PI/3 or 120 degrees.
tan(t)=-sqrt(3)
t=5PI/3 or 2PI/3
The period of tan is PI
The general solution is
t = 5PI/3+ n PI, where n is any integer
t = 2PI/3+ n PI, where n is any integer
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