$2\cos^2(3x)+5\cos(3x)−3<0$
For this question, I replaced $\cos(3x)$ by k which produced
$2k^2+5k−3<0$
Through factoring the left hand side I got
$(k+3)(2k-1)<0$
$0$ at $k=-3, 1/2$
Since the range of the cosine function is $[-1,1]$, $k=-3$ is extraneous
Equating $1/2$ to $\cos(3x)$, I got $x=\pi/9$
I know I need to find more values of $x$. However, I am not sure how to.
As $\cos3x+3\ge3-1>0,$
We need $\cos3x<1/2=\cos\pi/3$
Now the opposite $\cos3x\ge1/2\implies2n\pi-\pi/3\le3x\le2n\pi+\pi/3$ where $n$ is any integer