If sin(x) = 3/5 and x is in Quadrant II. in exact form find. Sin (x+pi/6)

I have no idea where to go about this

sin(x) = 3/5

cos(x) = √(1 - sin^2(x)) = √(1 - 9/25) = - 4/5 ( cos is negative in II quadrant )

sin(x + π/6) = sin(x) cos(π/6) + cos(x) sin(π/6)

= (3/5)(√3/2) + (-4/5)(1/2)

= (3/10)√3 - (2/5)

= 1/10 (3√3 - 4)

sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
sin(pi/6)=1/2
cos(pi/6)=sqrt(3)/2

sin (x+pi/6) = sin(x) cos(pi/6) + cos(x) sin(pi/6)
= (3/5) (sqrt(3)/2) + cos(x) (1/2)
cos(x) = +/- sqrt(1-sin^2(x)) = sqrt[1- (3/5)^2] = sqrt[ 1-9/25] = 4/5
Since cos is negative in quadrant II, cos(x) = -4/5

(3/5) (sqrt(3)/2) + cos(x) (1/2) = (3/5) (sqrt(3)/2) + (-4/5) (1/2)
= 3sqrt(3)/10 -4/10
= [3sqrt(3)-4]/10

sin(x + pi/6) = sin(x)*cos(pi/6) + cos(x)*sin(pi/6)

= (3/5)(sqrt(3)/2) - sqrt(1 - 9/25)(1/2)

= (3sqrt(3))/10 - 2/5