Question

Find the exact solution for the interval [0,2pi) -6sin4x=-12cos2x

Answer 1

LHS -6 sin 4x = (-6)2sin2xcos2x = -12sin2xcos2x = RHS = -12cos2x
-12 sin2x cos 2x + 12 cos2x =0
factor out -12 cos 2x ( sin2x -1) = 0
iff cos 2x =0 or sin2x -1 =0
a) in interval [0,2pi]; cos 2x =0 when 2x = pi/2 or 3pi /4
if 2x = pi /2---> x = pi/4
you do the same with 3pi/4 to get values of x
b) sin2x -1 = 0 ........... the same process.
at the end up. you must put all of them in conclusion to get final answer
I think it's quite clear. follow and if you need more. hopefully there is someone stops by and help. mine is done

Answer 2

how did you get to this -6 sin 4x = (-6)2sin2xcos2x. I understand it is the double angle formula but im confused.

Answer 3

sin 2x = 2 sinx cos x . yours is sin 2 (2x) . that 's it

Answer 4

but what happened to the 4x?

Answer 5

break it down to get 2(2x) and consider 2x as if it is one term