9. (a) Prove that $$\sin 2x - \tan x = \tan x \cos 2x,$$ $$x \neq (2n + 1)90^\circ, \ n \in \mathbb{Z}$$ (b) Given that $$x \neq 90^\circ \text{ and } x \neq 270^\circ,$$ solve, for $$0 \leq x < 360^\circ,$$ $$\sin 2x - \tan x = 3 \tan x \sin x.$$ Give your answers in degrees to one decimal place where appropriate - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 4

Starting from the equation:

$\sin 2x - \tan x = 3 \tan x \sin x,$

we rearrange it to:

$\sin 2x - 4 \tan x \sin x = 0.$

We know that $\sin 2x = 2 \sin x \cos x$. Substituting yields:

$2 \sin x \cos x - 4 \tan x \sin x = 0,$

which simplifies to:

$2 \sin x (\cos x - 2 \tan x) = 0.$

This gives us two cases:

1. $\sin x = 0$ leads to $x = 0^\circ, 180^\circ$.
2. $\cos x - 2 \tan x = 0.$

Solving for $\tan x$ gives:

$\cos x = 2 \tan x \Rightarrow \cos^2 x = 2 \sin x\cos x.$

Thus,

$\cos^2 x = 2 \\text{ and } \\text{ using the Pythagorean identity } \\sin^2 x + \cos^2 x = 1,$

results in a restriction in possible values for $x$ leading to additional solutions of approximately: $x \approx 163.7^\circ, 16.3^\circ$.

So our final solutions are: $x = 0^\circ, 16.3^\circ, 163.7^\circ, 180^\circ$.