Given is (1 - sin x) * tan x = sin x find all values for x between 0 and 2π

First of all, we'll substitute tan x= sin x/cos x.

(1 - sin x) * sin x/cos x = sin x

We'll open the paranthesis:

sin x - (sin x)^2=cos x*sin x

We'll write (sin x)^2=1- (cosx)^2

sin x - [1- (cosx)^2]=cos x*sin x

sin x - 1 + (cosx)^2 - cos x*sin x=0

We'll group the terms in this way: the first with the last  together, having sin x as common factor. We'll group the middle terms 1- (cos x)^2, seeing the group formed as a difference of squares, type a^2-b^2;

a^2-b^2=(a-b)(a+b)

sin x(1-cosx)-[1-(cosx)^2]=0

sin x(1-cosx)-(1-cosx)(1+cosx)=0

(1-cos x)(sinx-1-cosx)=0

A product of 2 factors is 0, when eather one or the other of the 2 factors is 0.

(1-cos x)=0

cos x=1, elementary equation

x=+/- arccos 1 + 2*k*pi

x=2*pi, when K=1

x=0, when k=0

The second factor, sinx-1-cosx=0

sinx-cosx=1, linear equation

This type of equation, due to the fact that the ratio between the coefficient of sin x and cos x is equal to the value 1, which it could be expressed as tan (pi/4)=1, could be solved with the help of the known angle. So the equation could be written in this way:

1*sinx-(1*cosx)=1

sin x - tan(pi/4)*cos x=1, but tan (pi/4)=sin(pi/4)/cos (pi/4)

sin x - sin(pi/4)/cos (pi/4)*cos x=1,

We'll have the same denominator all over, by multiplying  where necessary with the value cos(pi/4).

sin x*cos (pi/4) - sin(pi/4)*cos x=1*cos (pi/4)

sin [x-(pi/4)]=(sqrt 2)/2, instead of cos(pi/4)=(sqrt 2)/2

x-(pi/4)=(-1)^k*arcsin[(sqrt 2)/2]+k*pi

x=(-1)^k*(pi/4)+(pi/4)+k*pi

When k=0

x=(pi/4)+(pi/4)

x=2*(pi/4)

x=pi/2

When k=1

x=-(pi/4)+(pi/4)+pi

x=pi