
Trig Inequality
cos(4x) > cos(6x)
I've tried to let A = 2x, and solve it using duoble angles...
cos(2A) > cos(2A+A) ...then letting cos2AcosA  sin2AsinA replace cos(2A+A), etc. but it gets really messy and I get lost.
Can anyone please do this step by step?
Thanks in advance! :)

Hello, Sconts!
I found an approach, but it's just as bad as your expansion . . . LOL!
Quote:
cos(4x) .> .cos(6x)
I'll assume that the answer are between 0 and 2π.
I used a sumtoproduct identity: . cos A  cos B .= .2新in[(A+B)/2]新in[(AB)/2]
We have: .cos(4x)  cos(6x) .> .0
. . which becomes: .2新in(5x)新in(x) .> .0
And we have: .2新in(5x)新in(x) .> .0
There are two cases to consider:
. . [1] Both factors are positive.
. . [2] Both factors are negative.
[1] sin(x) > 0 .and .sin(5x) > 0
If sin(x) > 0, then: .0 < x < π
If sin(5x) > 0, then:
. . 0 < 5x < π . . → . . . .0 < x < π/5
. .2π < 5x < 3π . → . 2π/5 < x < 3π/5
. .4π < 5x < 5π . → . 4π/5 < x < π
Hence: .x ε (0, π/5) U (2π/5, 3π/5) U (4π/5, π)
[2] sin(x) < 0 .and .sin(5x) < 0
If sin(x) < 0, then: .π < x < 2π
If sin(5x) < 0, then:
. . 5π < 5x < 6π . . → . . . .π < x < 6π/5
. . 7π < 5x < 8π . . → . 7π/5 < x < 8π/5
. . 9π < 5x < 10π . → . 9π/5 < x < 2π
Hence: .x ε (π, 6π/5) U (7π/5, 8π/5) U (9π/5, π)
The solution is the union of all the intervals in blue.
But there must be a better way!

Thank you very much for the help, you saved me :D
I kept using half angle and cos(a+B) formulas and it got wayyy too messy, haha (I probably just didn't know how to simplify it, but okay :p)
Once again, thanks :)
Good method!