Solve the equation cos^2x=1-sinxcosx

We'll move the product sin x*cos x to the left side and
we'll shift (cos x)^2 to the right side:

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

The pythagorean identity
yields:

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

The equation will
become:

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

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

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

We'll factorize by sin
x:

sin x*(cos x - sin x) 
=0

We'll cancel each
factor:

sin x = 0 => x = (-1)^k*arcsin 0 +
2k*pi

x = 0 + 2k*pi

cos x -
sin x = 0

We'll divide by cos
x:

1 - tan x = 0

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

x = pi/4 +
k*pi

The sets of solutions of the equation
are: {0 + 2k*pi}U{pi/4 + k*pi}.