<PRE><B>Question 1007689</B>
4tan(theta)^2cos(theta)-2tan(theta)=0 
I found half the answer by dividing it all by 2tan(theta)^2 which turns into 
2cos(theta)=1 and then 
cos = 1/2. I am still missing another part.
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4tan(theta)^2cos(theta)-2tan(theta)=0 
Factor::
2tan(t)[2tan(t)cos(t) - 1) = 0
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2tan(t)[2sin(t)-1] = 0
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tan(t) = 0 or sin(t) = 1/2 
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t = 0 or pi OR t = pi/6 or (5/6)pi
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Cheers,
Stan H.
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