91Ó°ÊÓ

In the given problem we have to solve each equation on the interval $0≤\mathrm{θ}≤2\mathrm{π}$

localid="1646668211101" $\sin \left(3\mathrm{θ}+\frac{\mathrm{π}}{18}\right)=1$

So, we know that $3\mathrm{θ}+\frac{\mathrm{π}}{18}$must equal $\frac{\mathrm{π}}{2}$.

To find these solutions, write the general formula that gives all the solutions.

localid="1646668266127" $3\mathrm{θ}+\frac{\mathrm{Ï€}}{18}=\frac{\mathrm{Ï€}}{2}+2\mathrm{Ï€²Ô}$

Subtract $\frac{\mathrm{Ï€}}{18}$on both side

localid="1646668365473" role="math" $$\begin{gathered} 3\mathrm{θ}+\frac{\mathrm{Ï€}}{18}-\frac{\mathrm{Ï€}}{18}=\frac{\mathrm{Ï€}}{2}+2\mathrm{Ï€²Ô}-\frac{\mathrm{Ï€}}{18} \\ 3\mathrm{θ}=\frac{\mathrm{Ï€}}{2}·\frac{9}{9}+2\mathrm{Ï€²Ô}·\frac{18}{18}-\frac{\mathrm{Ï€}}{18} \\ 3\mathrm{θ}=\frac{9\mathrm{Ï€}+36\mathrm{Ï€²Ô}-\mathrm{Ï€}}{18} \\ 3\mathrm{θ}=\frac{8\mathrm{Ï€}+36\mathrm{Ï€²Ô}}{18} \end{gathered}$$

Divide by 3 on both side

localid="1646668392976" $$\begin{gathered} \frac{3}{3}\mathrm{θ}=\frac{8\mathrm{Ï€}+36\mathrm{Ï€²Ô}}{18}·\frac{1}{3} \\ \mathrm{θ}=\frac{8\mathrm{Ï€}+36\mathrm{Ï€²Ô}}{54} \end{gathered}$$

So the value of given function in interval $[0,2\mathrm{Ï€})$is

localid="1646668731677" $$\begin{gathered} \mathrm{θ}=\frac{8\mathrm{π}+36\mathrm{π}×0}{54}\mathrm{θ}=\frac{8\mathrm{π}+36\mathrm{π}×1}{54}\mathrm{θ}=\frac{8\mathrm{π}+36\mathrm{π}×2}{54} \\ \mathrm{θ}=\frac{8\mathrm{π}}{54}\mathrm{θ}=\frac{8\mathrm{π}+36\mathrm{π}}{54}\mathrm{θ}=\frac{8\mathrm{π}+72\mathrm{π}}{54} \\ \mathrm{θ}=\frac{8\mathrm{π}}{54}\mathrm{θ}=\frac{44\mathrm{π}}{54}\mathrm{θ}=\frac{80\mathrm{π}}{54} \end{gathered}$$

So the solution set is $\left\{\frac{8\mathrm{Ï€}}{54},\frac{44\mathrm{Ï€}}{54},\frac{80\mathrm{Ï€}}{54}\right\}$.