What is the solution to cos^2x - sin^2x + sin x = 0.

Since there are 2 terms that contain the fuction sine,
we'll replace (cos x)^2, by the difference 1 - (sin x)^2 (from Pythagorean
identity).

The equation will
become:

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

We'll multiply by -1 and we'll re-arrange the
terms:

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

We'll replace sin x by another variable
t:

2t^2 - t - 1 = 0

We'll
solve the quadratic:

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

t1 = (1+3)/4

t1 =
1

t2 = -1/2

We'll put sin x =
t1 => sin x = 1

x = k*pi/2 +
k*pi

sin x = t2

sin x =
-1/2

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

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