We have to find a for the identity 2sin^2a-sina-1=0 to be true.

Now 2* (sin a)^2 - sin a - 1 =0

let y = sin a

=> 2y^2 - y - 1 =0

=> 2y^2 - 2y + y - 1 =0

=> 2y( y - 1) + 1(y-1) = 0

=> (2y + 1)( y-1) = 0

=> y = -1/2 and y = 1

As y = sin x

x = arc sin (-1/2)

=>x = -30 + k*360 degrees

x = arc sin 1

=> x = 90 + k*360 degrees

**Therefore x is -30 + k*360 degrees and 90 + k*360 degrees.**

We'll apply substitution technique and we'll note sin a = t.

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

2t^2 - t - 1 = 0

Since it is a quadratic equation, we'll apply 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 a = t1.

sin a = 1

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

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

Now, we'll put sin a = t2.

sin a = -1/2

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

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

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

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