Trig help!!

[allh5]

It's solve sin4x=2cos2x [0,2 pie)

I subtracted the 2cox2x over so everything equals zero so I'm not sure where to go from there.

 

[soroban]

Duplicate post

 

[nyc_function]

Trig Equation

This post should be in the Geometry and Trig forum.

 

[Deleted member 4993]

Duplicate post

Nope.... the other one was killing the student!!!!

 

[raghvendrasingh]

It's solve sin4x=2cos2x [0,2 pie)

I subtracted the 2cox2x over so everything equals zero so I'm not sure where to go from there.

ok.....

sin4x=2cos2x
2sin2xcos2x-2cos2x=0
2cos2x(sin2x-1)=0

so cos2x=0 or sin2x=1

now if cos2x=0 then x=(2n+1)pie/4 where n is an integer and and n belongs to [0,3].

now if sin2x=1 then x=mpie/2 +(-1)^m pie/4 where m is an integer and m belongs to [0,3]
.