Solve $$\csc^2 2x - \cot 2x = 1$$ for $0 \leq x \leq 180^\circ.$ - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 2

Rearranging gives:

$\cot^2 2x - \cot 2x - 1 = 0$.

This is a quadratic equation in terms of (\cot 2x).

We can factor or use the quadratic formula:

$\cot 2x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}$,

which simplifies to:

$\cot 2x = \frac{1 \pm \sqrt{5}}{2}$.

Calculating gives two principal values:

1. (\cot 2x = 2.618... o 2x = \tan^{-1} \left(\frac{1}{2.618...}\right)) which approximates to (22.5^\circ).

2. (\cot 2x = -0.618... o 2x = 180^\circ - \tan^{-1} \left(\frac{1}{0.618...}\right)) which gives another solution near (112.5^\circ).

Calculating further within the range yields (x = 22.5^\circ, 112.5^\circ).

Within the interval (0 \leq x \leq 180^\circ), the complete solutions are:

$x = 22.5^\circ, 112.5^\circ, 45^\circ, 135^\circ$.