document.write( "Question 551337: sin (2x) = cos (2x) between (0, 2pi) \n" ); document.write( "
Algebra.Com's Answer #359654 by Alan3354(69443)\"\" \"About 
You can put this solution on YOUR website!
sin (2x) = cos (2x) between (0, 2pi)
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\n" ); document.write( "sin^2 = cos^2 = 1 - sin^2
\n" ); document.write( "2sin^2 = 1
\n" ); document.write( "sin^2(2x) = 1/2
\n" ); document.write( "\"sin%282x%29+=+sqrt%282%29%2F2\"
\n" ); document.write( "\"sin%282x%29+=+-sqrt%282%29%2F2\"
\n" ); document.write( "2x = pi/4, 3pi/4, 5pi/4, 7pi/4, 9pi/4, 11pi/4, 13pi/4, 15pi/4
\n" ); document.write( "x = pi/8, 3pi/8, 5pi/8, 7pi/8, 9pi/8, 11pi/8, 13pi/8, 15pi,8
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\n" ); document.write( "x = pi/8 + n*pi/4, n = 0 to 7
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