sinx = 1+ cos^2 x

First we will use the trigonometric identities to simplify and solve for x.

We know that:

cos^2 x = 1- sin^2 x.

==> sinx = 1+ ( 1-sin^2 x)

==> sin(x) = 1+1 + sin^2 x

==> sin(x) = 2- sin^2 x

==> sin^2 x + sinx - 2 = 0

==> Let y = sinx.

==> y^2 + y - 2 = 0

==> (y+2) ( y-1) = 0

==> y1= -2 ==> sinx = -2 ( not valid).

==> y2= 1 ==> sinx = 1 ==> x = pi/2 + 2npi + 3pi/2+ 2npi.

Then, there are unlimited solutions for x.

**==> x= { pi/2+ 2npi, 3pi/2 + 2npi} n = 0, 1, 2, 3, ....**

We have to solve sin x = 1 + (cos x)^2

Let sin x = y.

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

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

=> 1- y^2

Now we can write sin x = 1 + (cos x)^2 as

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

=> y = 2 - y^2

=> y^2 + y - 2 = 0

=> y^2 + 2y - y - 2 =0

=> y(y + 2) - 1(y+2) = 0

=> (y-1)(y+2) = 0

=> y = 1 or y = -2

As y = sin x

sin x = 1 ( we can't have sin x = -2)

As sin x = 1

**x = pi/2 + 2*n*pi**

We'll write (cos x)^2 with respect to (sin x)^2, from the fundamental formula of trigonomtery:

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

We'll substitute (1) in the given equation:

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

We'll combine like terms and we'll move all terms to the right side:

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

We'll substitute sin x by t:

t^2 + t - 2 = 0

We'll apply the quadratic formula:

t1 = [-1+sqrt(1 + 8)]/2

t1 = (-1+3)/2

t1 = 1

t2 = (-1-3)/2

t2 = -2

But t1 = sin x => sin x = 1

x = (-1)^k*arcsin (1) + k*pi

**x = (-1)^k*(pi/2) + k*pi**

**We'll reject the second solution for t since -1 =< sin x =< 1.**

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