Solve the trigonometric equation sqrt(2)sin(x/3)+1=0

Enter each answer in the form θ + 2πk, with θ ≥ 0 and θ as small as possible

There are apparently 2 solutions -

my steps:

sqrt(2)sin(x/3)+1=0

substitute y for x/3

sin(y) = -1/sqrt2 <-- both positive and negative quadrants

sin^-1(y) = sin^-1(sqrt2/2)

y = pi/4

x/3 = pi/4

x = 3pi/4.....

I know this is wrong, but what is the right way to solve?

Thank you all!

sqrt(2) * sin(x/3) + 1 = 0
sqrt(2) * sin(x/3) = -1
sin(x/3) = -1 / sqrt(2)
sin(x/3) = -sqrt(2)/2
x/3 = 3pi/4 + 2pi * k , 5pi/4 + 2pi * k
x/3 = (pi/4) * (3 + 8k) , (pi/4) * (5 + 8k)
x = (3pi/4) * (3 + 8k) , (3pi/4) * (5 + 8k)
x = (9pi/4) + 6pi * k , (15pi/4) * 6pi * k