What is the solution of equation sin^2x+sinx*cosx=1?

We'll shift (sin x)^2 to the right
side:

sin x*cos x = 1 - (sin
x)^2

But, from Pythagorean identity, we'll
have:

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

The equation will
become:

sin x*cos x = (scos
x)^2

We'll subtract (cos x)^2 both
sides:

sin x*cos x - (cos x)^2 =
0

We'll factorize by cos
x:

cos x*(sin x - cos x) 
=0

We'll cancel each
factor:

cos x = 0 => x = +/-arccos 0 +
2k*pi

x = +/-(pi/2)
+ 2k*pi

sin x - cos x =
0

We'll divide by cos x:

tan x
- 1 = 0

tan x = 1 => x = arctan 1 +
k*pi

x = pi/4 +
k*pi

The solutions of the equation belong to
the reunion of the following sets: {+/-(pi/2) + 2k*pi}U{pi/4 +
k*pi}.