document.write( "Question 785500: solve 3sin(2x+40)=2cos(2x+40) where 0
Algebra.Com's Answer #477586 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! solve 3sin(2x+40)=2cos(2x+40) where 0 < x <360 \n" ); document.write( "sub t for 2x+40 (saves typing) \n" ); document.write( "----- \n" ); document.write( "3sin(t) = 2cos(t) \n" ); document.write( "9sin^2 = 4cos^2 = 4(1 - sin^2) \n" ); document.write( "9sin^2 = 4 - 4sin^2 \n" ); document.write( "13sin^2 = 4 \n" ); document.write( "sin^2 = 4/13 \n" ); document.write( "sin(2x+40) = +2sqrt(13)/13 \n" ); document.write( "x = 356.845 degs \n" ); document.write( "------------- \n" ); document.write( "sin(2x+40) = -2sqrt(13)/13 \n" ); document.write( "=========================== \n" ); document.write( "x = 323.155 degs \n" ); document.write( " |