Solve, for $0 \leq \theta \leq 180^{\circ}$, $$2\cot^2(30^{\circ}) = 7\csc(30^{\circ}) - 5$$ Give your answers in degrees to 1 decimal place. - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 6

Now, simplifying both sides:

$\frac{2}{3} = 14 - 5$

This leads to:
$\frac{2}{3} = 9,$ which is incorrect, indicating that we should derive solution for $heta$.

We can reorder and form an equation:

$2\cot^2(\theta) = 7\csc(\theta) - 5$

Substituting again: $2\cot^2(\theta) = 14 - 5 = 9$ This will yield to finding $heta$.

After reorganizing, $\cot^2(\theta) = \frac{9}{2}$ This leads to $\theta = \arccot\left(\frac{3}{\sqrt{2}}\right)$
Calculate principal solution:

Using calculator, find: $\theta \approx 53.5^{\circ}.$ Also, check the domain and the possible secondary solution: $\theta = 180^{\circ} - 53.5^{\circ} \approx 126.5^{\circ}.$