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## -sin^2=2cosx-2

solve the following equation...I am confused...

what i have so far:

2cosx-2+sin^2=0

....

- Sinx^2 +2 cosX -2

Remember

SinX^2 + Cos X^2 =1, then SinX^2 = 1 - cosx^2

Substitute :

CosX^2 -1 +2cosX -2 = 0

Cos X^2 +2CosX -3 =0

(CosX +3)( CosX -1 ) =0

CosX =1 is the only answer.
X = 0  , X= 2n(360)

Hi Mary,

Remember the Pythagorean identity, sin^2(x) + cos^2(x) = 1.  If you use this to replace sin^2 with an expression containing cos^2, you can factor the expression.

So you had sin^2 (x) + 2 cos (x) - 2 = 0  and you know by the pythagorean identity that

sin^2 (x) = 1 - cos^2 (x), so replace sin^2 in your equation:

(1 - cos^2 (x) ) + 2 cos (x) - 2 = 0     combine 1-2 = -1 ...

- cos^2 (x) + 2 cos (x) - 1 = 0  take the opposite of each term in the equation...

cos^2 (x) - 2 cos (x) + 1 = 0  and this can be factored.

It might be easier to look at if you let y = cos (x)

y^2 -2y +1 = (y - 1)(y - 1)  so replace y with cos (x) and you have

(cos (x) - 1 )(cox (x) - 1 ) = 0  by the zero product property, cos (x) - 1 = 0,

so cos (x) = 1, and the angle that has a cosine of 1 is 0,

so x = 0 (plus any multiple of 360 degrees or 2 pi radians)