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

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We have to solve (sin x)^2 + sinx*cosx = 1

We know that (sin x)^2 + (cos x)^2 = 1

As (cos x)^2 is replaced by sin x*cos x in the equation, we can equate the two:

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

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

=> (cos x)(cos x - sin x) = 0

cos x = 0 for x = 90 + n*360

and cos x = sin x for x = 45 + n*360

**The values of x that satisfy this are x = 90 + n*360 degrees and x = 45 + n*360 degrees.**

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}.**