sin(3x) + cos(3x) = -1.    (1)   (It is the original equation)

Square its both sides. You will get

 = .   (2)

From the other side, there is an identity

 == .   (3)

Comparing (2) and (3), you get

2*sin(3x)*cos(3x) = 0,   or  sin(3x)*cos(3x) = 0.    (4)

Equation (4) splits in two independent equations

1)  sin(3x) = 0  --->  3x = ,   k = 0. +/-1. +/-2, . . . 
                 --->   x = ,  k = 0, +/-1. +/-2, . . .       (5)

2)  cos(3x) = 0  --->  3x = ,   k = 0. +/-1. +/-2, . . . 
                 --->   x = ,  k = 0. +/-1. +/-2, . . .   (6)

Now we should check which of the found values (5), (6) satisfy the original equation.

Of the set (5), all x satisfy sin(3x) = 0. Hence, only those of (5) satisfy the original equation where cos(3x) = -1.
They are  3x = ,    n = 0, +/-1. +/-2, . . . ,  or 

           x = ,  n = 0, +/-1. +/-2, . . . ,           (5').

Of the set (6), all x satisfy cos(3x) = 0. Hence, only those of (6) satisfy the original equation where sin(3x) = -1.
They are  3x = ,   n = 0, +/-1. +/-2, . . . ,  or 

           x = ,  n = 0, +/-1. +/-2, . . . ,           (6'). 

Answer.  The union of the sets (5') and (6') is the solution of the original equation.