You need to write the equation in terms of x such that:

2sin x*cos x + 2 sin x - cos x - 1 = 0

You may form two groups of terms such that:

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

Factoring out cos x in group (2sin x*cos x - cos x) yields:

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

You need to factor out (2sin x - 1) such that:

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

You should solve equations 2 sin x - 1 = 0 and cos x + 1 = 0 such that:

2 sin x - 1 = 0 => 2 sin x = 1 => sin x = 1/2 => x=(-1)^n*sin^(-1) (1/2)+npi

x = (-1)^n(pi/6) + npi

cos x = -1 => x = pi + 2npi

**Hence, the solutions to equation are x = (-1)^n(pi/6) + npi and x = pi + 2npi.**

We'll apply the double angle identity for the first term:

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

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

sin 2a = 2sin x*cos x

We'll re-write the equation:

2sin x*cos x + 2sinx - cosx - 1 = 0

We'll factorize by 2sin x the first 2 terms:

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

We'll factorize by (cos x + 1):

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

We'll cancel each factor:

cos x + 1 = 0

We'll shift 1 to the right:

cos x = -1

x = arccos (-1)

x = pi

2sin x - 1 = 0

We'll add 1 both sides:

2sin x = 1

sin x = 1/2

x = arcsin (1/2)

x = pi/6

x = 5pi/6

The values of the angle x are: {pi/6 ; 5pi/6 ; pi}.