Solve the equation cos^2x=1-sinxcosx

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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}.**

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