How do I find the x-intercepts of y= 2sin3x + √3

How do I find the x-intercepts of y= 2sin3x + √3 for 0 ≤ x ≤ 2π?

Thanks!

x intercepts so y =0
so
2sin3x+√3=0
2sin3x=-√3
sin3x=-√3/2
you do sine inverse to find the angle and you get -60 degrees
3x=-60+360k
or 3x=240+360k
so
x=-20+120k
or
x=80+120k
and you start substituting for values of K
When K=-1 This is rejected because you get negative answers
when K=0 you get an accepted answer of 80 for the second equation( the first equation has a negative answer so it is rejected )
when k=1 You get two accepted answers 100 and 200
when K=2 you get two accepted answers which are 220 and 320
When k=3 the answers are so big so they are rejected
so the answers are {80,100,200,220,320) degrees
convert them to radians
{4pi/9,5pi/9,11pi/9 and 16pi/9)
So the key to answer this questions is to know that sine is :
theta+360k
and
180-theta+360k
where k can be any integer

solve for y=0

0= 2sin3x + √3
2sin3x=-√3
sin3x=-√3/2

so it must be either
3x=4π/3

or

3x=5π/3

then x must be

x=4π/9 or x=5π/9