8. (a) Prove that $$2\cot 2x + \tan x \equiv \cot x$$ $$x \neq \frac{n\pi}{2}, n \in \mathbb{Z}$$ (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 9 - 2016 - Paper 3

To prove the equation, we start by using the double angle identity:

$\cot 2x = \frac{\cos 2x}{\sin 2x} = \frac{2\cos^2 x - 1}{2\sin x \cos x}$.

So, substituting this in gives us:

$2\cot 2x = 2\cdot \frac{2\cos^2 x - 1}{2\sin x \cos x} = \frac{2\cos^2 x - 1}{\sin x \cos x}$.

Next, we rewrite (\tan x = \frac{\sin x}{\cos x}), thus, the left side becomes:

$2\cot 2x + \tan x = \frac{2\cos^2 x - 1}{\sin x \cos x} + \frac{\sin x}{\cos x} = \frac{2\cos^2 x - 1 + \sin^2 x}{\sin x \cos x}$.

Using the Pythagorean identity (\sin^2 x + \cos^2 x = 1):

$2\cot 2x + \tan x = \frac{2\cos^2 x + \sin^2 x - 1}{\sin x \cos x} = \frac{\cos^2 x + 1 - 1}{\sin x \cos x} = \frac{\cot x}{\sin x \cos x}.$

Thus, we have established that:

$2\cot 2x + \tan x = \cot x,$

completing the proof.