solve this

ln(2x+1) < sin4x

Find the range of x

uhm, right...

care to elaborate?

OP here. No one who knows the solution? :/

ln (2x+1) < sin4x
R = sin 4x = 2sin2xcos2x= 4 sinx cosx * (cos^2 x - sin^2 x)
= 4 sinx cos x * (1 - 2 sin^2 x) = 4 sinx cosx - 8sin^3 x or consider right side as ln with sin 4x belongs to <-1;1> and 2x+1 > 0, here x > -1/2, I'd love to help more and find a solution but im off for a beer, cheers :P

No idea how to solve that. Mathematica says "The equations appear to involve the variables to be solved for in an essentially non-algebraic way."

Solving numerically the range appears to be
[-0.5 ... -0.304329] ⋃ [0 ... 0.569456]

Yeah, it's most likely -1/2 I will try to graph the two and deduce it. As transcendental relations don't really have algebraic solutions.

>>7
It's -1/2 because ln(2x+1) is undefined anywhere farther left.

>>6
It's strange that Mathematica includes -.5 as a possible value, since not both functions are defined there.

op again :) thanks a lot everyone. but... how do I reach  x > -1/2 ?

>>9
from: ln(2x+1). 2x+1 > 0 => x > -1/2

>>9
>>6
My bad, that should be an open interval ][ for both parts