Solve the equation `tan x + tan(x+pi/3) +tan(x+(2pi)/3)=3 ` for `0 ltx lt2pi`

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`tanx + tan(x+pi/3) + tan(x+(2pi)/3)=3`

To simplify, use the sum of angle identity for tangent. Formula is:

`tan (A+B) = (tanA+tanB)/(1-tanAtanB)`

So we have,

`tanx + (tanx+tan(pi/3))/(1-tanxtan((pi)/3))) + (tanx + tan((2pi)/3))/[1-tanx tan((2pi)/3)]=3`

`tanx + [tanx +sqrt3]/[1-sqrt3tanx] + [tanx-sqrt3]/[1+sqrt3tanx] = 3`

Multiply both sides by the LCD `(1-sqrt3tanx)(1+sqrt3tanx)` .

`tanx(1-3tan^2x)+(tanx+sqrt3)(1+sqrt3tanx)+(tanx-sqrt3)(1-sqrt3tanx) `


`tanx-3tan^3x + 4tanx+sqrt3tan^2x+sqrt3 + 4tanx-sqrt3tan^2x-sqrt3 `


`9tanx-3tan^3x = 3-9tan^2x`

To simplify further, divide both sides by 3.

`3tanx - tan^3x= 1 - 3tan^2x`


Group the terms so that it would lead to a common factor between the two groups.

`(tan^3x+1) -(3tan^2x+3tanx) = 0`

Factor each group. For the first group, use the factor of sum of two cubes which is `a^3+b^3 = (a+b)(a^2-ab+b^2)` . 

And for the second group, factor out 3tanx.


Then factor out GCF.



Set each factor to zeroa and solve for x.

`tan x + 1 = 0`


`x= (3pi)/4` and `(7pi)/4`

For the second factor, use the quadratic formula. So, let z=tanx.


`z = [-b+-sqrt(b^2-(4ac))]/(2a) = [-(-4)+-sqrt[(-4)^2-4*1*1]]/(2*1)=2+-sqrt3`

For each value of z, solve angle x.

`z=2+sqrt3` ,         `2+sqrt3=tanx`

                                      `x = (5pi)/12`  and `(17pi)/12`

`z=2-sqrt3` ,             `2-sqrt3=tanx `             

                                       `x=pi/12` and `(13pi)/12`

Hence, the values of x in the equation

`tanx + tan(x+pi/3) + tan(x+(2pi)/3)=3` for the interval `0<x<2pi` are

`pi/12` ,  `(5pi)/12` ,  `(3pi)/4` ,  `(13pi)/12` ,  `(17pi)/12`  and  `(7pi)/4` .

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