Determine the x angle if 2sin^2x+sin(-x)-1=0

For the beginning, we remark that one term has as
argument the opposite variable, -x. Since the sine function is odd, we'll write the
term:

sin(-x) = - sin x

We'll
re-write the given expression:

2 (sin x)^2 - sin x - 1 =
0

We'll substitute sin x =
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 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 solutions of the equation are:
{(-1)^k*(pi/2) + k*pi}U{(-1)^(k+1)*(pi/6) +
k*pi}.