document.write( "Question 1028991: Find all the values of x in the interval [0,2pi] such that the gradient of the curve y=12sinx-5cosx is 13/2. (using differentiation) \n" ); document.write( "
Algebra.Com's Answer #644079 by Alan3354(69443) You can put this solution on YOUR website! Find all the values of x in the interval [0,2pi] such that the gradient of the curve y=12sinx-5cosx is 13/2. \n" ); document.write( "----------- \n" ); document.write( "y' = 12cos(x) + 5sin(x) = 13/2 \n" ); document.write( "12*sqrt(1 - sin^2) = -5sin + 13/2 \n" ); document.write( "144 - 144sin^2 = 25sin^2 - 65sin + 169/4 \n" ); document.write( "169sin^2 - 65sin - 407/4 = 0 \n" ); document.write( "------- \n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "================== \n" ); document.write( "sin(x) = 0.991715757339482 \n" ); document.write( "x =~ 82.62 degs, 97.38 degs \n" ); document.write( "=========== \n" ); document.write( "sin(x) = -0.607100372724097 \n" ); document.write( "x =~ 217.38 degs, 322.62 degs \n" ); document.write( "================= \n" ); document.write( "Convert those to radians. \n" ); document.write( " \n" ); document.write( " |