<PRE><B>Question 719291</B>
Solve the following for B between 0 and 360 degrees: 
(1+4 sin B)(3-2 sin B)=2
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Let x = sin(B)
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(1+4x)(3-2x) = 2
3 + 10x - 8x^2 = 2
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8x^2 -10x - 1 = 0
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x = [10 +- sqrt(100-4*8*-1)]/16
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x = [10 +- sqrt(132)]/16
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Usable solution:
x = -0.0931
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Solve for &quot;B&quot;
sin(B) = -0.0931
B = sin^-1(-0.0931) = 354.66 degrees or B = 185.34 degrees
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Cheers,
Stan H.
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