Question 810211
Use the double angle formula
(1) sin(2*x) = 2*sin(x)*cos(x)
and the givens
(2) cos(x) = -1/3 and
(3) sin(x) < 0
Use {{{sin^2(x) + cos^2(x) = 1}}} to get
(4) {{{sin^2(x) + (-1/3)^2 = 1}}} or
(5) {{{sin^2(x)  = 1 -1/9}}} or
(6) {{{sin^2(x) = 8/9}}} or
(7) {{{sin(x) = - sqrt(8)/3}}}
where we select the negative value in (7) because of the given condition (3).
Now put sin(x) and cos(x) into (1) to get
(8) sin(2*x) = 2*(- sqrt(8)/3)*(-1/3) or
(9){{{ sin(2*x) = 2*sqrt(8)/9}}} or
(10) {{{sin(2*x) = 4*sqrt(2)/9}}}
Let's use numerical values to check the answer.
From (2) we get
(11) x = arccos(-1/3) or
(12) x = 109.47+
Then we get
(13) sin(x) = -0.9428
and
(14) sin(2x) = sin(218.94) or
(15) sin(2x) = 0.6285
Now 
Is (sin(2*x) = 2*sin(x)*cos(x))?
Is (0.6285 = 2*(-09428)*(-1/3))?
Is (0.6285 = 0.6285)? Yes
Answer:{{{sin(2*x) = 4*sqrt(2)/9}}}