You need to use the identity `sin(-alpha) = -sin alpha` , such that:

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

You need to group the terms such that:

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

Converting the difference of squares` sin^2 x - 1` into a product, yields:

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

Factoring out` sin x` yields:

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

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

`(sin x - 1)(sin x + 1 + sin x) = 0 => {(sin x - 1 = 0),(2sin x + 1 = 0):} => {(sin x = 1 ),(sin x = -1/2):} => {(x = (-1)^n*sin^(-1) 1 + n*pi),(x = (-1)^n*sin^(-1)(-1/2) + n*pi):} => {(x = (-1)^n*(pi/2) + n*pi),(x = (-1)^(n+1)*(pi/6) + n*pi):}`

**Hence, evaluating the solutions to trigonometric equation yields **`x = (-1)^n*(pi/2) + n*pi, x = (-1)^(n+1)*(pi/6) + n*pi.`

By definition, the sine function is odd,therefore, we'll re-write the term sin(-x):

sin(-x) = - sin x

We'll re-write the given expression:

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

We'll replace sin x by t.

We'll re-write the equation using the new variable t:

2t^2 - t - 1 = 0

We'll use the quadratic formula:

t1 = [1 + sqrt(1 + 8)]/4

t1 = (1+3)/4

t1 = 1

t2 = (1-3)/4

t2 = -1/2

We'll put sin x = t1.

sin x = 1

x = (-1)^k*arc sin 1 + k*pi

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

Now, we'll put sin x = t2.

sin x = -1/2

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

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

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

The required solutions of the equation are: {(-1)^k*(pi/2) + k*pi}U{(-1)^(k+1)*(pi/6) + k*pi}.