# What is x if cos x = 1 + sin^2 x?

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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**

cos x = 1 + sin^2 x.

To find x.

We subtract 1+ sin^2x from both side of the given equation:

cosx-sin^2x- 1 = 0.

We substitute for sin^2x = 1-cos^2x.

cosx-(1-cos^2x)-1 = 0.

We rearrange the equation:

cos^2x+cosx-2 = 0.

(cosx-1)(cosx+2) = 0

cosx-1 = 0 or cosx+2 = 0

cosx-1 = 0 gives cosx = 1.

x = 2npi, where n = 0,1,2,....