You need to solve the following trigonometric equation by factoring, such that:

`2sin^2 x - 3sinx + 1 = 0 => 2sin^2 x - 2sinx - sin x + 1 = 0`

You need to group the terms such that:

`(2sin^2 x - 2sinx) - (sin x - 1) = 0`

Factoring out `2 sin x` in the first group yields:

`2sin x(sin x - 1) - (sin x - 1) = 0`

Factoring out` (sin x - 1)` yields:

`(sin x - 1)(2sin x - 1) = 0 => {(sin x - 1 = 0),(2sin x - 1 = 0):}`

`{(sin x = 1),(sin x = 1/2):} => {(x = +-pi/2 + n*pi),(x = +-pi/6 + npi):}`

**Hence, evaluating solutions to trigonometric equation, using factorization, yields **`x = +-pi/2 + n*pi, x = +-pi/6 + npi. `

We'll shift all terms to the left side:

2(sin x)^2- 3sinx + 1 = 0

Now, we'll apply substitution technique to solve the equation.

Let sin x = t and we'll re-write the equation in t:

2t^2 - 3t + 1 = 0

Since it is a quadratic, we'll apply the quadratic formula:

t1 = {-(-3) + sqrt[(-3)^2 - 4*2*1]}/2*2

t1 = [3+sqrt(9-8)]/4

t1 = (3+1)/4

t1 = 1

t2 = (3-1)/4

t2 = 1/2

Now, we'll put sin x = t1.

sin x = 1

Since it is an elementary equation, we'll apply the formula:

sin x = a

x = (-1)^k* arcsin a + 2k*pi

In our case, a = 1:

x = (-1)^k* arcsin 1 + 2k*pi

x = (-1)^k*(pi/2) + 2k*pi

x = pi/2

Now, we'll put sin x = t2

sin x = 1/2

x = (-1)^k* arcsin 1/2 + 2k*pi

x = (-1)^k* (pi/6) + 2k*pi

x = pi/6

x = pi - pi/6

x = 5pi/6

**The solutions of the equation are:{ pi/6 ; pi/2 ; 5pi/6 }.**