Given that 2sinx+1=2sin^2x/cosx+sinx/cosx what is x.

We'll re-write the given equation and we'll multiply by
cos x, having all terms to one side:

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

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

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

We'll factorize by (2 sin x +
1):

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

We'll put 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 put the next factor as
zero:

sin x - cos x = 0

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

tan x - 1 = 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, in the interval[0 , 2pi], are: {pi/4 ; 5pi/4 ; 7pi/6 ;
11pi/6}.