Given that $\sin^{2}\theta + \cos^{2}\theta \equiv 1$, show that $1 + \cot^{2}\theta \equiv \csc^{2}\theta$ - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 5

1. Rewrite the equation:
   [ 2 \cot^{2}\theta - 9 \csc\theta - 3 = 0 ]

2. Substitute $\csc\theta$ and $\cot\theta$ in terms of sine and cosine functions:
   [ \cot\theta = \frac{\cos\theta}{\sin\theta} \text{ and } \csc\theta = \frac{1}{\sin\theta} ]

3. This leads to:
   [ 2 \left( \frac{\cos^{2}\theta}{\sin^{2}\theta} \right) - 9 \left( \frac{1}{\sin\theta} \right) - 3 = 0 ]

4. Rearranging terms gives the quadratic:
   [ 2\csc^{2}\theta - 9\csc\theta - 6 = 0 ]

5. Applying the quadratic formula $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ where $a=2$, $b=-9$, and $c=-6$: [ \csc\theta = \frac{9 \pm \sqrt{(-9)^{2} - 4(2)(-6)}}{2(2)} ]
   which simplifies to: [ \csc\theta = \frac{9 \pm \sqrt{81 + 48}}{4} = \frac{9 \pm \sqrt{129}}{4} \approx 3.5 \text{ or } -0.5 ]

6. Therefore, the valid solutions for $\theta$ (not within the range for $\csc\theta$) will be from the positive value, leading to:
   [ \theta \approx 11.5^{\circ}, 168.5^{\circ} ]

Thus, the solutions for $\theta$ up to 1 decimal place are $11.5^{\circ}$ and $168.5^{\circ}$.