Solve the equation: 2 sin 2x + 3*(sin x+cos x) = -2.

We'll solve this equation
algebraically.

We'll move all terms to one
side:

2sin2x + 3(sinx+cosx) + 2 =
0

We'll substitute sin 2x = 2sin x*cos
x

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

But (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:

t^2 + 4t + 5 = 0

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

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

The only solution of
the equation is: x = -pi/4 +
kpi.