The points of inflection of a function are given by the zeros of the second derivative. The first derivative of our function is:
![f^(\prime)(x)=2(-\csc(3x)/(2)\cot(3x)/(2))\cdot(3)/(2)=-3(\csc(3x)/(2)\cot(3x)/(2))](https://img.qammunity.org/2023/formulas/mathematics/college/b6oa995j322bqbj1qjlp7btxa03shuhok9.png)
Using the product rule, the second derivative of our function is:
![\begin{gathered} f^(\prime)^(\prime)(x)=-3\lbrack(\csc(3x)/(2))^(\prime)\cdot(\cot(3x)/(2))+(\csc(3x)/(2))\cdot(\cot(3x)/(2))^(\prime)\rbrack \\ \\ =(9)/(2)(\csc(3x)/(2)\cot^2(3x)/(2)+\csc^3(3x)/(2)) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/p51dqrvlo1wl2jbdu5tsehuq9pivzniiay.png)
Then, the zeros of the second derivative are the solutions for the following equation:
![(9)/(2)(\csc(3x)/(2)\ctg^2(3x)/(2)+\csc^3(3x)/(2))=0](https://img.qammunity.org/2023/formulas/mathematics/college/pdjgn16rp48t021kemzlejpw3tp9m49xis.png)
Simplifying this expression, we have:
![\begin{gathered} (9)/(2)(\csc(3x)/(2)\ctg^2(3x)/(2)+\csc^3(3x)/(2))=0 \\ \\ \csc(3x)/(2)\ctg^2(3x)/(2)+\csc^3(3x)/(2)=0 \\ \\ \ctg^2((3x)/(2))+\csc^2((3x)/(2))=0 \\ \\ \cos^2((3x)/(2))+1=0 \\ \\ \cos^2((3x)/(2))=-1\implies\\exists x\in\mathbb{R} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mzfd9fv2h531p125cqssssicjkxaf0v4vh.png)
Our function has no inflection points.