# What is x if cos^2(x/2)-2cos^2x=(3/2)*square root2(1+cosx)+2sin^2x? We have to solve cos^2(x/2)-2cos^2x=(3/2)*square root2(1+cosx)+2sin^2x for x.

Rewriting the equation in a simpler form:

(cos x/2)^2 - 2 ( cos x)^2 = (3/2)sqrt 2 ( 1 + cos x) + 2 (sin x)^2

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

=> (cos x/2)^2 = (3/2)(sqrt 2)(1...

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We have to solve cos^2(x/2)-2cos^2x=(3/2)*square root2(1+cosx)+2sin^2x for x.

Rewriting the equation in a simpler form:

(cos x/2)^2 - 2 ( cos x)^2 = (3/2)sqrt 2 ( 1 + cos x) + 2 (sin x)^2

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

=> (cos x/2)^2 = (3/2)(sqrt 2)(1 + cos x) + 2

=> (cos x/2)^2 = (3/sqrt 2)(1 + 2(cos x/2)^2 - 1) + 2

=> (cos x/2)^2 = 3*sqrt 2(cos x/2)^2 + 2

=> (cos x/2)^2 - 3*sqrt 2(cos x/2)^2 = 2

=> (cos x/2)^2 (1 - 3*sqrt 2) = 2

=> (cos x/2)^2 = 2/(1 - 3*sqrt 2)

=> cos x/2 = sqrt (2/(1 - 3*sqrt 2))

=> x/2 = arc cos ( sqrt (2/(1 - 3*sqrt 2)))

Therefore x = 2*arc cos ( sqrt (2/(1 - 3*sqrt 2))) + 2*n*pi

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