Question 1063200
 I need to solve this equation for 0 < x< 2pi
-2tan(-2x + 4pi/3) = 2sqrt3 
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tan(-2x+(4/3)pi) = -sqrt(3)
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Take the arctan of both sides to get:
-2x+(4/3)pi = (2/3)pi or (5/3)pi
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-2x = (-2/3)pi or (1/3)pi
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x = (1/3)pi or (-2/3)pi
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Cheers,
Stan H.