Question

I need to solve:

sin(4x)=3cos(2x)

Answer 1

@hartnn maybe you could help me now?

Answer 2

write sin 4x = 2 sin 2x cos 2x

Answer 3

we know that\[\sin 2 \theta = 2\sin \theta \cos \theta\]

Answer 4

yeah, I did it

Answer 5

2 sin 2x cos 2x do i need to rewrite it again, using sin(2x) and cos(2x) again?

Answer 6

so
\[2\sin 2x\cos 2x=3\cos 2x\]
\[\cos2x(2\sin 2x-3)=0\]
\[\cos2x=0\] or \[2\sin 2x-3=0\]

Answer 7

didnt get :/ can you explain?

Answer 8

which step you didn't get ?
he factored out cos 2x after subtracting 3cos2x from both sides..

Answer 9

oh, understand now:) Thank you a lot:)

Answer 10

didnt understand how -3 appeared in the breckets, but I do now:) THANK YOU A LOT!

Answer 11

2sin2x−3=0 doesnt have sollutions, right?

Answer 12

correct.

Answer 13

just solve cos 2x=0

Answer 14

thanks, hartnn!:)

Answer 15

welcome :)