document.write( "Question 551337: sin (2x) = cos (2x) between (0, 2pi) \n" ); document.write( "
Algebra.Com's Answer #359654 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! sin (2x) = cos (2x) between (0, 2pi) \n" ); document.write( "------- \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( " \n" ); document.write( " \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 \n" ); document.write( "----------- \n" ); document.write( "x = pi/8 + n*pi/4, n = 0 to 7 \n" ); document.write( " \n" ); document.write( " |