The original equation to be solved is

`2*tan^2(2x) =6+tan(2x) `

We make the substitution `tan(2x) =y ` and obtain a second degree equation in `y ` :

`2y^2 =6+y <=> 2y^2-y-6 =0`

The solutions are

`y_1 =(1+sqrt(1+48)]/4 =2 ` and `y_2 =[1-sqrt(1+48)]/4 =-3/2 `

We have therefore

`tan(2x) =2 and...

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The original equation to be solved is

`2*tan^2(2x) =6+tan(2x) `

We make the substitution `tan(2x) =y ` and obtain a second degree equation in `y ` :

`2y^2 =6+y <=> 2y^2-y-6 =0`

The solutions are

`y_1 =(1+sqrt(1+48)]/4 =2 ` and `y_2 =[1-sqrt(1+48)]/4 =-3/2 `

We have therefore

`tan(2x) =2 and tan(2x) =-1.5 `

The graph of function `f(x) =tan(x) ` is periodic with the periodicity `pi ` ,

therefore the graph of function `g(x) = tan(2x) ` is periodic with the periodicity `pi/2 ` (see attached graph below).

For the first equation `tan(2x) =2 ` we have the general solutions `2x=arctan(2) +n*pi/2 `, where `n in ZZ ` .

In the interval `(2x) in [-2pi,pi] ` these solutions are:

`2x =arctan(2) -2*pi ` , for `n=-4 ` . Thus `x =1/2*arctan(2) -pi `

`2x =arctan(2)-3*pi/2 ` , for `n=-3 ` . Thus `x =1/2*arctan(2) -3*pi/4 `

`2x=arctan(2) -pi ` , for `n=-2 ` . Thus `x =1/2*arctan(2) -pi/2 `

`2x =arctan(2)-pi/2 ` , for `n=-1 ` . Thus `x =1/2*arctan(2) -pi/4 `

`2x =arctan(2) ` , for `n=0 ` . Thus `x =1/2*arctan(2) `

`2x =arctan(2)+pi/2 ` , for `n=+1 ` . Thus `x = 1/2*arctan(2) +pi/4 `

Here above `1/2*arctan(2) =0.554 rad `

For the second equation `tan(2x) =-1.5 ` the general solution is of the form `2x =arctan(-1.5) +n*pi/2 ` , where `n ` is integer.

For this particular value of `tan(2x) =-1.5 ` there are other 6 values of `2x ` in the interval `[-2pi,pi] ` .

These happen for the values of `n=-3,-2,-1,0,1,2 ` .

**Thus there are 12 values of `x ` for which the above equation is satisfied in the interval `(2x) in [-2pi.pi] ` **

**These values are `2x =arctan(2) +n*pi/2 ` with `n=-4,-3,..,+1 ` and `2x =arctan(-1.5)+n*pi/2 ` with `n =-3,-2,...,+2 ` .**