document.write( "Question 597421: Solve the equation on the interval [0,2pi)\r
\n" ); document.write( "\n" ); document.write( "2 cos^2 x + sin x -2 = 0
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Algebra.Com's Answer #378169 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Solve the equation on the interval [0,2pi)
\n" ); document.write( "2 cos^2 x + sin x -2 = 0
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\n" ); document.write( "2(1-sin^2) + sin -2 = 0
\n" ); document.write( "----
\n" ); document.write( "2-2sin^2+sin -2 = 0
\n" ); document.write( "2sin^2 - sin = 0
\n" ); document.write( "sin(x)(2sin(x)-1) = 0
\n" ); document.write( "sin(x) = 0 or sin(x) = 1/2
\n" ); document.write( "x = 0 or pi or 2pi or pi/6 or (5/6)pi
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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