2cos²(3x) + 5cos(3x) - 3 < 0

The left factors just as 2u²+5u-3 = (u+3)(2u-1)

[cos(3x)+3][2cos(3x)-1] < 0

We find the critical values by finding the zeros of the left side

The first factor has no zeros since no cosine can be -3
The second factor has zeros when cos(3x) = 

This is when    and when 
                        
                       
                         



From the graph of y = 2cos²(3x) + 5cos(3x) - 3 we see that the graph is negative
(below the x-axis) in all open intervals
, when n is any integer.       

Edwin