3 Answers

sin(4x)  cos(2x) = 0
2sin(2x)cos(2x)  cos(2x) = 0
cos(2x)(2sin(2x)  1) = 0
cos(2x) = 0
in which case
2x = pi/2, 3pi/2, 5pi/2
x = pi/4, 3pi/4, 5pi/4
x = (pi/4 + npi/2) where n is an integer
or
2sin(2x)  1 = 0
2sin(2x) = 1
sin(2x) = 1/2
2x = pi/6, 5pi/6, 13pi/6, 17pi/6, ...
x = pi/12, 5pi/12, 13pi/12, 17pi/12, ...

sin4x + cos2x = 0 2sin2x . cos2x + cos2x =0 cos 2x ( 2sin 2x +a million) =0 cos 2x =0 2x=nkp + p/2 x =nkp/2 + p/4 2sin 2x +a million =0 sin2x =a million/2 2x =2kp p/6 x=kp p/12 2x =2kp +7p/6 x=kp + 7p/12

i dont know if this is right but i would make it 2 seperate equations like so:
sin4x=0 and cos2x=0
and if the equation is correct then x should be the same in both meaning u have x