ii) Solve $3{\csc }^{2}2\theta +5\cot 2\theta =31(0^{\circ }<\theta <{180}^{\circ })$

Step 1 ：First, we rewrite the given equation using the reciprocal identities for cosecant and cotangent. The reciprocal of cosecant is sine, and the reciprocal of cotangent is tangent. So, the equation becomes $3{\sin }^{-2}2\theta +5\tan 2\theta =31$.

Step 2 ：Next, we use the Pythagorean identity $1+{\tan }^2\theta ={\sec }^2\theta$ to rewrite the equation in terms of sine and cosine. This gives us $3{\sin }^{-2}2\theta +5\sin 2\theta {\cos }^{-1}2\theta =31$.

Step 3 ：Now, we multiply through by ${\cos }^{2}2\theta$ to clear the denominators. This gives us $3{\sin }^{2}2\theta {\cos }^{2}2\theta +5\sin 2\theta =31{\cos }^{2}2\theta$.

Step 4 ：We can simplify this equation by using the double-angle identities $\sin 2\theta =2\sin \theta \cos \theta$ and $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$. This gives us $3(2\sin \theta \cos \theta {)}^2+10\sin \theta \cos \theta =31({\cos }^2\theta -{\sin }^2\theta )$.

Step 5 ：Expanding and simplifying, we get $12{\sin }^2\theta {\cos }^2\theta +10\sin \theta \cos \theta =31{\cos }^2\theta -31{\sin }^2\theta$.

Step 6 ：This is a quadratic equation in $\sin \theta \cos \theta$. We can solve it by factoring or using the quadratic formula. The solutions are $\sin \theta \cos \theta =\frac{-10\pm \sqrt{100-4(12)(-31)}}{24}$.

Step 7 ：Solving for $\sin \theta \cos \theta$, we get two solutions: $\sin \theta \cos \theta =\frac{-10+\sqrt{1372}}{24}$ and $\sin \theta \cos \theta =\frac{-10-\sqrt{1372}}{24}$.

Step 8 ：We can find the solutions for $\theta$ by taking the inverse sine of these values. However, we must remember that $\sin \theta \cos \theta$ is equal to $\frac{1}{2}\sin 2\theta$, so we need to divide our solutions by 2 before taking the inverse sine.

Step 9 ：Doing this, we find that the solutions for $\theta$ are $\theta =\frac{1}{2}{\sin }^{-1}\left(\frac{-10+\sqrt{1372}}{24}\right)$ and $\theta =\frac{1}{2}{\sin }^{-1}\left(\frac{-10-\sqrt{1372}}{24}\right)$.

Step 10 ：However, we must check these solutions to make sure they are in the interval $0^{\circ }<\theta <{180}^{\circ }$. If they are not, we must discard them.

Step 11 ：After checking, we find that the only solution in the given interval is $\boxed{{\displaystyle \theta =\frac{1}{2}{\sin }^{-1}\left(\frac{-10+\sqrt{1372}}{24}\right)}}$.