# How do you solve -2tan(x)=x? I am confused as to how to use arctan when I have two x's

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`-2tan(x)=x`

If you try to solve this analytically, it is almost impossible to arrive at a solution.

`tan(x) = -x/2`

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

But you can find the solution in an approximate analytical solution.

lt's write tan(x) as an taylor expansion.

Taylor expansion

Taylor expansion of f(x) around x =a is given by,

`f(x) = f(a)+(f'(a))/(1!)(x-a)+(f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+....`

let's expand tan(x) until 3rd order around x =0. So a =0

if `tan(x) = f(x)` ` f(0) = 0`

`f'(x) = sec^2(x)` ` f'(0) = 1`

`f''(x) = 2sec(x)sec(x)tan(x) = 2sec^2(x)tan(x) = 2f'(x)f(x)`

`f''(0) = 0`

`f'''(x) = 2f'(x)f'(x)+2f''(x)f(x) = 2(f'(x))^2+2f(x)f''(x)`

`f'''(0) = 2*1^2+0 = 2`

Therefore Taylor expansion of tan(x) is,

`tan(x) = 0+1/(1!)x+0+2/(3!)x^3`

`tan(x) = x + 2/(1*2*3)x^3`

`tan(x) = x + x^3/3`

now using this to sovle our problem,

-2tan(x) =x

`-2(x+x^3/3) = x`

`2/3x^3+3x = 0`

`x(2/3x^2+3) = 0`

this gives, either `x = 0` or `(2/3x^2+3) = 0`

But x should be a real number therefore, `2/3x^2+3 != 0`

The only answer is x = 0.

**The answer is x =0**