## anonymous 4 years ago sin^2(3x)=1−cos6x , can someone explain the process?

1. anonymous

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

2. anonymous

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

3. anonymous

Jimmyrep had the right idea...

4. anonymous

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

5. anonymous

Oh .

6. anonymous

Not so bright is correct...to sub

7. anonymous

$\sin^23x=0$ $3x=n \pi$

8. anonymous

yes i think notsobright is correct

9. anonymous

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

10. anonymous

yup

11. anonymous

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

12. anonymous

Thanks!

13. anonymous

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

14. anonymous

Alright .

15. anonymous

notsobright is correct

16. anonymous

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...

17. anonymous

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

18. anonymous

Got it , thanks again !