Question 1036596
Hi there,
You have not listed limits
so I am working this out
for 0<=&#952;=<2&#960;
2 sin(3&#952;) + 1 = 0
2 sin(3&#952;) = -1
sin(3&#952;) = -1/2
(3&#952;) = &#960;/6, 5&#960;/6
Because you have (3&#952;) you must 
add 2&#960; to each of your values twice:-
(3&#952;) = &#960;/6, 5&#960;/6,13&#960;/6,17&#960;/6
25&#960;/6 and 29&#960;/6.
Now you divide all the values by 3.
&#960; = &#960;/18, 5&#960;/18,13&#960;/18,17&#960;/18
25&#960;/18 and 29&#960;/18.
Hope this helps :-)