What is x if 2*sin x + 1 = tan x + 2*sin x * tan x, if x lies in [0; 2*pi] ?

We'll use the fact that the function tangent is a
ratio:

 tan x = sin x/cos
x

We'll re-write the given equation moving all terms to one
side:

2 sin x + 1 -  2 sin x (sin x/cos x) - sin x/cos x =
0

We'll multiply by cos
x:

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

We'll factorize the first 2 terms by cos x  and the last
2 terms by - sin x :

cos x(2 sin x + 1) - sin x(2 sin x +
1) = 0

We'll factorize by 2 sin x +
1:

(2 sin x + 1)(cos x - sin x) =
0

We'll set the first factor as
zero:

2 sin x + 1 = 0

We'll
subtract 1;

2sinx = -1

sin x =
-1/2

x = arcsin (-1/2)

The
sine function is negative in the 3rd and 4th quadrants:

x =
pi + pi/6

x = 7pi/6 (3rd
qudrant)

x = 2pi -
pi/6

x = 11pi/6 (4th
qudrant)

We'll set the other factor as
zero:

cos x - sin x = 0

This
is an homogeneous equation and we'll divide it by cos x:

1
- tan x = 0

tan x = 1

The
function tangent is positive in the 1st and the 3rd
qudrants:

x = arctan
1

x = pi/4 (1st
quadrant)

x = pi+
pi/4

x = 5pi/4 (3rd
qudrant)

The complete set
of solutions of the equation, over the interval[0 , 2pi], are: {pi/4 ; 5pi/4 ; 7pi/6 ;
11pi/6}.