document.write( "Question 921574: can someone please help me with this question ? A water wheel rotates through the angle theta,the water level L behind the wheel changes according to the equation: L =2-2sintheta-4cos^2theta where l is measured in inches. determine the values of theta for which the water level is zero. find the exact values. do not use a claculator \n" ); document.write( "
Algebra.Com's Answer #559075 by Alan3354(69443)\"\" \"About 
You can put this solution on YOUR website!
L =2 - 2sin(t) - 4cos^2(t) = 0
\n" ); document.write( "1 - sin(t) - 2cos^2(t) = 0
\n" ); document.write( "1 - sin(t) - 2(1 - sin^2(t)) = 0
\n" ); document.write( "2sin^2 - sin - 1 = 0
\n" ); document.write( "(2sin + 1)*(sin - 1) = 0
\n" ); document.write( "sin = 1 --> theta = 90 + n*180 degs, n = 0,1,2,3...
\n" ); document.write( "----
\n" ); document.write( "sin = -1/2 --> theta = 210 + n*360 degs, n = 0,1,2,3...
\n" ); document.write( "sin = -1/2 --> theta = 330 + n*360 degs, n = 0,1,2,3...
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