Solve the equation 2-3cosx+2square root((1-cos4x)/2)=3sinx

First, we'll move the term 3cosx to the right
side:

2 + 2sqrt((1-cos4x)/2) = 3sinx +
3cosx

We'll apply the half angle identity for the term
sqrt((1-cos4x)/2)= sin 2x.

2 + 2sin 2x = 3(sin x + cos
x)

We'll apply the double angle identity for sin
2x:

sin 2x = 2 sin x*cos
x

We'll solve this equation
algebraically.

We'll move all terms to one
side:

2sin2x - 3(sinx+cosx) + 2 =
0

2*2sin x*cos x - 3(sinx+cosx) + 2 =
0

We'll note sin x + cos x =
y.

We'll raise to square and we'll
get:

(sin x + cos x)^2 =
y^2

(sin x)^2 + (cos x)^2 + 2sin x*cos x =
y^2

From Pythagorean
identity:

 (sin x)^2 + (cos x)^2 =
1

1 + 2sin x*cos x = y^2

2sin
x*cos x = y^2 - 1

We'll re-write the given equation in
y:

2*(y^2 - 1) - 3(y) + 2 =
0

We'll remove the
brackets:

2y^2 - 2 - 3y + 2 =
0

We'll eliminate like
terms:

2y^2 - 3y = 0

We'll
factorize by y:

y(2y - 3) =
0

We'll put y = 0

But y = sin
x + cos x:

sin x + cos x =
0

We'll divide by cos x:

tan x
+ 1 = 0

tan x = -1

x = -pi/4 +
kpi

2y + 3 = 0

2y =
3

y = 3/2

But the range of
values of y is [-2;2].

Maximum of the sum: sin x + cos x =
1 + 1 = 2

Minimumof the sum: sin x + cos x = -1-1 =
-2

We'll work with
substitution:

sin x =
2t/(1+t^2)

cos t =
(1-t^2)/(1+t^2)

2t/(1+t^2) + (1-t^2)/(1+t^2) =
3/2

4t + 2 - 2t^2 = 3 +
3t^2

We'll move all terms to one
side:

-5t^2 + 4t -1 =
0

t1 =
[-4+sqrt(16-20)]/2

Since delta is negative, the equation
has no real solutions.

The equation will have
only one solution: x = -pi/4 +
kpi.