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