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

use the property that (sin x)^2 = 1 - (cos x)^2

cos x = 1 + sin^2 x

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

=> (cos x)^2 + cos x - 2 = 0

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

=> cos x( cos x + 2) - 1(cos x + 2) = 0

=> (cos x - 1)(cos x + 2) = 0

=> cos x = 1 and cos x = -2

but cos x cannot be -2, so we can eliminate that.

We have cos x = 1

=> x = 0 + n*360

**The solution is n*360 degrees.**

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

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

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

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

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

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

We'll substitute cos 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 = cos x => cos x = 1

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

**x = 0 + 2k*pi**

**x = 2kpi**

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