How do you solve `Tan(x+pi)=-2sin(x+pi)`?

Remember that

`tan(x+pi)=[tanx+tanpi]/[1-tanx*tanpi]`

But `tanpi=0` hence

`tan(x+pi)=tanx`

Also

`sin(x+pi)=sinx*cospi+cosx*sin(pi)=-sinx`

So now we have that

`tan(x+pi)=-2sin(x+pi)=>tanx=2sinx=> sinx(1/cosx-2)=0`

Hence we have that `sinx=0=>x=k*pi` where `k` an integer

Also `cosx=1/2=>x=2*n*pi+-pi/3` where `n` an integer

But we have to exclude the values for which `cosx=0`

hence

`x!=m*pi-pi/2` where `m` is an integer.

Hence the solutions are `{k*pi,2*n*pi+-pi/3}` except values at

`{m*pi-pi/2}`

Call `(x + pi) = a`, we get:
`sin a/(cos a) = -2sin a`
sin a = -2sin a.cos a
`sin a(1 + 2cos a) = 0`
a. sin a = 0 --> a = 0, and a = pi, and a = 2pi.
Therefor, `x = -pi`, and x = 0, and `x = pi`-->
x = 0 and `x = pi` for `(0, 2pi)`
b. 1 + 2cos a = 0 -->`cos a = -1/2` -->arc `a = +- (2pi)/3`.
1. `(x + pi) = (2pi)/3` --> `x = (2pi)/3 - pi = - pi/3`, or `x = (5pi)/3`.
2. `(x + pi) = -(2pi)/3`
`x = -(2pi)/3 - pi = -(5pi)/3,` or `x = pi/3`
Answer for (0, 2pi):
`0, pi/3, pi, and (5pi)/3.`
Check.
`x = pi/3` --> `tan (pi/3 + pi) = tan pi/3 = sqrt3`;
`- 2sin (pi/3 + pi) = 2sin pi/3 = (2sqrt3)/2 = sqrt3`. OK