document.write( "Question 1025601:  root3cosx+sinx-2=0
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| Algebra.Com's Answer #640889 by Alan3354(69443)     You can put this solution on YOUR website! root3cosx+sinx-2=0 \n" ); document.write( "---- \n" ); document.write( "If you mean sqrt(3)*cos(x) + sin(x) - 2 = 0: \n" ); document.write( "sqrt(3)*cos(x) = -sin(x) + 2 \n" ); document.write( "Square \n" ); document.write( "3cos^2 = sin^2 - 4sin + 4 \n" ); document.write( "3(1-sin^2) = sin^2 - 4sin + 4 \n" ); document.write( "3-3sin^2 = sin^2 - 4sin + 4 \n" ); document.write( "4sin^2 - 4sin + 1 = 0 \n" ); document.write( "(2sin - 1)^2 = 0 \n" ); document.write( "sin(x) = 1/2 \n" ); document.write( "etc \n" ); document.write( " \n" ); document.write( " |