Find the angle a for the identity 2sin^2a-sina-1=0 to be true.

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}.