document.write( "Question 635251: solve over the interval [0,2pi)
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document.write( "2sin^2(x)+cos(x)-1=0 \n" );
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Algebra.Com's Answer #400213 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! solve over the interval [0,2pi) \n" ); document.write( "2sin^2(x)+cos(x)-1=0 \n" ); document.write( "---- \n" ); document.write( "2(1-cos^2(x))+cos(x)-1 = 0 \n" ); document.write( "2 - 2cos^2 + cos - 1 = 0 \n" ); document.write( "---- \n" ); document.write( "2cos^2 - cos -1 = 0 \n" ); document.write( "Factor: \n" ); document.write( "(2cos+1)(cos-1) = 0 \n" ); document.write( "cos(x) = -1/2 or cos(x) = 1 \n" ); document.write( "x = (2/3)pi or x = (4/3)pi or x = 0 \n" ); document.write( "============ \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |