What is the angle x if the identity sin2x-(cos 2x)/2 -1/2=cosx-2cosx*tanx?

To determine the angle x, we'll have to solve the
equation.

We'll write the first term of the equation as the
function sine of a double angle.

We'll apply the formula
for the double angle:

sin 2a = sin (a+a)=sina*cosa +
sina*cosa=2sina*cosa

We'll replace 2a by 2x and we'll
get:

sin 2a = 2sin x*cos
x

We'll re-write the
equation:

2sin x*cos x - (cos 2x)/2 -1/2 =
cosx-2cosx*tanx

We'll factorize by -1/2 the last 2 terms
from the left side:

2sin x*cos x - (1-cos 2x)/2 =
cosx-2cosx*tanx

But (1-cos 2x)/2 = (cos
x)^2

2sin x*cos x - (cos x)^2 = cosx-2cosx*sinx/cos
x

We'll simplify and we'll
get:

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

We'll factorize by cos x to the left
side:

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

We'll move all the terms to the left
side:

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

We'll factorize by (2sin x - cos
x):

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

We'll set each factor as
zero:

cos x + 1 = 0

We'll add
-1 both sides:

cos x = -1

x =
arccos (-1)

x = pi

2sin x -
cos x = 0

We'll divide by cos
x:

2tan x -  1 = 0

tan x =
1/2

x = arctan
(1/2)

The angle x has the following values: {
pi ; arctan (1/2) }.