Show that 2 tan x - cot x = 5 cosec x may be written in the form $a \cos^2 x + b \cos x + c = 0$ stating the values of the constants $a$, $b$ and $c$ - Edexcel - A-Level Maths Pure - Question 4 - 2014 - Paper 6

From the earlier manipulation, we have established: $3 \cos^2 x + 5 \cos x - 2 = 0$

Using the quadratic formula: $\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting $a = 3$, $b = 5$, and $c = -2$:

$\cos x = \frac{-5 \pm \sqrt{5^2 - 4(3)(-2)}}{2(3)} = \frac{-5 \pm \sqrt{25 + 24}}{6}$

Calculating further: $\cos x = \frac{-5 \pm 7}{6}$

This results in two cases:

1. (\cos x = \frac{2}{6} = \frac{1}{3})
2. (\cos x = \frac{-12}{6} = -2) (reject as not valid)

For (\cos x = \frac{1}{3}), we find: $x = \cos^{-1}\left(\frac{1}{3}\right)$ Calculating further, we find two values for $x$ in the range $0 \leq x < 2\pi$: $x \approx 1.231 \text{ and } x \approx 5.050$

Thus, to 3 significant figures, $x \approx 1.23$ and $x \approx 5.05$.

Starting with: $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$

Combining these yields: $\tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}$

Recognizing that: $\sin 2\theta = 2 \sin \theta \cos \theta$

Thus, we can rewrite our expression: $\tan \theta + \cot \theta = \frac{2}{\sin 2\theta} = \lambda \csc 2\theta$

Identifying the constant: $\lambda = 2$.