8. (a) Prove that $$2 \cot 2x + \tan x \equiv \cot x \\ x + \frac{n \pi}{2}, n \in \mathbb{Z}$$ (4) (b) Hence, or otherwise, solve, for $$-\pi < x < \pi,$$ $$6 \cot 2x + 3 \tan x = \csc^2 x - 2$$ Give your answers to 3 decimal places - Edexcel - A-Level Maths Pure - Question 1 - 2016 - Paper 3

1. Start by rewriting the equation:
   $6 \cot 2x + 3 \tan x = \csc^2 x - 2$

2. Utilize the identity $\csc^2 x = 1 + \cot^2 x$, so the equation becomes:
   $6 \cot 2x + 3 \tan x = 1 + \cot^2 x - 2$

3. Rearranging gives:
   $\cot^2 x + 6 \cot 2x + 3 \tan x + 1 = 0$

4. Consider $\cot 2x = \frac{\cos 2x}{\sin 2x}$ and substitute back, applying the double angle formulas.
   Simplify and rearrange terms to form a quadratic in terms of $\tan x$:
   $\tan^2 x = \frac{3}{\sqrt{3}}$

5. Solving gives:
   $\tan x = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

Calculate the angles within the interval:

• For $x \approx 0.294$ (1st quadrant)
• For $x \approx -2.848$ (3rd quadrant)
• For $x \approx -1.865$ (2nd quadrant)

Finally, the answers in three decimal places are:
$x \approx 0.294, -2.848, -1.865$