Question

sin^2(3x)=1−cos6x , can someone explain the process?

Answer 1

sin^2(3x) = 1 - cos^2(3x)
might be useful here

Answer 2

no
start with cos 6x = 1 - 2sin^2(3x)
1 - cos 6x = 2sin^2(3x)
- are you sure you have the correct expression

Answer 3

Jimmyrep had the right idea...

Answer 4

\[1-\cos(6x)=2\sin^23x\]

Answer 5

Oh .

Answer 6

Not so bright is correct...to sub

Answer 7

\[\sin^23x=0\]
\[3x=n \pi\]

Answer 8

yes i think notsobright is correct

Answer 9

cos(2u)=1-2sin^2(u)

Answer 10

yup

Answer 11

since u = 3x
then cos2(3x)= 1-2sin^2(3x)

Answer 12

Thanks!

Answer 13

this is what jimmyrep had to start with...not what is given...make sure you copied it down correctly.

Answer 14

Alright .

Answer 15

notsobright is correct

Answer 16

I first thought he/she were trying to verify that they were equal....
but as notsobright and kingkos has indicated...it appears that you were solving for x in which case not so bright was correct...

Answer 17

values in which the sin^2(3x)=0...or when 3x=npi as was stated.

sorry for the confusion.

Answer 18

Got it , thanks again !