Question

2cos4x+2sin^2 2x+4cos^2 2x=6

Answer 1

2cos(4x)+2(1-cos^2(2x))+4(cos(4x)-1/2)=6
2cos(4x)+2(1-((cos(4x)-1)/2))+4((cos(4x)-1)/2) = 6

Answer 2

it's enough ??

Answer 3

well i got somehow cos4x=3 , but i think it should be between [-1,1]

Answer 4

oh i got cos 4x= 5/3 but i still think it's wrong!

Answer 5

i think the right value is x=0

Answer 6

well give me a minute or two :)

Answer 7

i find
3cos(4x)=3

Answer 8

i use
cos^2(x)+sin^2(x)=1
and
cos(4x) = 2cos^2(x)-1

Answer 9

but can you show me step after that second you told me, because cos4x and stuff like that is totally new for me!

Answer 10

ok

Answer 11

2cos(4x)+2(1-cos^2(2x))+4cos(2x) = 6
2cos(4x) +2-2cos^2(2x)+4cos(2x) =6
2cos(4x)+2cos^2(2x)=4
---------------
because of
cos(4x) = 2cos^2(x)-1
then 2cos^2(2x)=cos(4x)+1
-----------------------
2cos(4x)+cos(4x)+1=4
3cos(4x)=3
x= kpi/2

Answer 12

thanks a lot !

Answer 13

i hope
i helped you

Answer 14

well that helps alot :)

Answer 15

btw. in start there was ...4cos^2(2x)=6
and in your start there is ....4cos(2x)=6, why is that

Answer 16

just i forget the square 2

Answer 17

but , it stays right (i forget to write it just)