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

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 $\boxed{\theta = \frac{1}{2} \sin^{-1} \left(\frac{-10 + \sqrt{1372}}{24}\right)}$.