# What is x if cos^2(x/2)-2cos^2x=(3/2)*square root2(1+cosx)+2sin^2x?

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### 2 Answers

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**

The first step is to move all terms to the left side side:

(cos x/2)^2 - 2(cos x)^2 - (3/2)*square root2(1+cosx) - 2(sin x)^2 = 0

We recognize the formula for the half angle:

square root[(1+cosx)/2] = (cos x/2)

We'll re-group the terms:

(cos x/2)^2 - 2[(sin x)^2 + (cos x)^2] - [3*****2(cos x/2)]/2= 0

We'll write the fundamental formula of trigonometry;

[(sin x)^2 + (cos x)^2] = 1

The equation will become:

(cos x/2)^2 - 2- 3(cos x/2) = 0

We'll substitute cos x/2 = t

t^2 - 3t - 2 = 0

We'll apply quadratic formula:

t1 = [-3 + sqrt(9 + 8)]/2

t1 = (-3 + sqrt17)/2

t2 = (-3 - sqrt17)/2

But cos x/2 = t

cos x/2 = t1

cos x/2 = (-3 + sqrt17)/2

x = +/-2arccos[(-3 + sqrt17)/2] + 2kpi

cos x/2 = t2

cos x/2 = (-3 - sqrt17)/2, impossible since cos a > = - 1.

**The only solution of the equation is: {+/-2arccos[(-3 + sqrt17)/2] + 2kpi}.**