document.write( "Question 1025601: root3cosx+sinx-2=0
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Algebra.Com's Answer #640889 by Alan3354(69443)\"\" \"About 
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root3cosx+sinx-2=0
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\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
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