Question 670975
An equation is given. (Enter your answers as a comma-separated list. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)
&#8722; 2/sqr3 cos 3&#952; = 1
**
[0,2&#960;]
cos 3&#952;=1/(-2/&#8730;3)=-&#8730;3/2
3&#952;=7&#960;/6 and 11&#960;/6 in quadrants III and IV where cos<0
&#952;=7&#960;/18 and 11&#960;/18