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

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}.