<PRE><B>Question 858018</B>
Find the value of x in degree 5 (sin ^2)x + 12 sin x + 4 = 0
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5sin^2(x) + 12sin(x) + 4 = 0
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Factor:
5sin^2(x) + 10sin(x)+2sin(x) + 4 = 0
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5sin(x)[sin(x)+2][2((sin(x)+2)] = 0
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(sin(x)+2)(5sin(x)+2] = 0
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sin(x) = -2 (extraneous)
sin(x) = -2/5 = -0.4
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Ans: x = arcsin(-0.4) = -23.58 degrees or 180+23.58 degrees
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Cheers,
Stan H. </PRE>