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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

Q: What are all the exact values for which tan t equals sqrt3?

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y = 2x is defined everywhere (for all values of x)

True

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y = 2x is defined everywhere (for all values of x)

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that would be limited to 3 and -3 for values of x

True

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