What x satisfies the identity -2(sin2x+1)/3=sinx+cosx ?

First, we'll move all terms to one
side:

-2(sin2x+1)/3-(sinx+cosx) =
0

We'll multiply by -3 the
equation:

2sin2x +  2 + 3(sinx+cosx) =
0

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

 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

From Pythagorean identity, we'll
have:

 (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

This is a homogenous trigonometric equation in sinx and
cos x. We'll divide the equation by cos x:

sin x/cos x + 1
= 0

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.

Therefore, x will have
only one value that satisfies the identity: x = -pi/4 +
kpi.