sin4xcos2xcos4xsin2xsquare root 2sinxSolutionsin4xcos2xcos4x

\"sin4xcos2x-cos4xsin2x=(square root 2)sinx\"

Solution

sin4xcos2x-cos4xsin2x=(square root 2)sinx

sin(4x-2x) = (square root 2)sinx

sin(2x) = (square root 2)sinx

2sinx*cos(x) = (square root 2)sinx

sinx = 0 is one of the roots x = 2n

2cos(x) = (square root 2)

cos(x) = 1/(square root 2)

x = 2n + /4

so roots are 2n + /4 and 2n