You should use the following formulas such that:

`sin 2x = (2tan x)/(1 + tan^2 x)`

`cos 2x= (1 - tan^2 x)/(1 + tan^2 x)`

You need to substitute `t` for `tan x` such that:

`6*(1 - t^2)/(1 + t^2) + (2k-1)*(2t)/(1 + t^2)= 8`

`6 - 6t^2 + 4kt - 2t - 8 - 8t^2 = 0`

`-14t^2 + 2t*(2k-1) - 2 = 0`

Dividing by -2 yields:

`7t^2 - t(2k-1) + 1 = 0`

You should use quadratic formula to evaluate t such that:

`t_(1,2) = (2k-1 +- sqrt((2k-1)^2 - 28))/14`

You should notice that the radicand`(2k-1)^2 - 28` needs to be positive for`t_1` and `t_2` to be real values such that:

`(2k-1)^2 - 28 >= 0`

Solving the attached equation yields:

`(2k-1)^2 - 28 = 0 => (2k-1)^2 = 28 => 2k-1 = +-sqrt28`

`2k-1 = +-2sqrt7 => 2k = 1+-2sqrt7 => k = (1+-2sqrt7)/2`

Notice that the expression is positive for `k in (-oo, (1-2sqrt7)/2)U((1+2sqrt7)/2, +oo).`

**Hence, there exists solution to the given trigonometric equation if `k in (-oo, (1-2sqrt7)/2)U((1+2sqrt7)/2, +oo).` **