How do I solve 2sin4x cosx = 0 in the interval [0, 2pi)?

Answer 1

well, either

sin4x=0
or
cosx=0

so, 4x = 0,pi,2pi,3pi,4pi,...
meaning x = n*pi/4 for n=0 to 7

or, x = pi/2 or 3pi/2
but those are already covered by sin4x=0

see

http://www.wolframalpha.com/input/?i=+2sin4x+cosx

Answer 2

every time sin 4x = 0
and every time cos x = 0

sin 4 x = 0 when
x = 0, pi/4 , pi/2 , 3 pi/4, pi , 5 pi/4.... 8 pi/4 which is 2 pi

cos x = 0 when x = pi/2 , 3 pi/2 but we already have those two points