3. (i) (a) Show that 2 tan x - cot x = 5 cosec x may be written in the form a cos² 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

Using the quadratic equation derived previously:

$3 \cos^2 x + 5 \cos x - 2 = 0,$

we can apply the quadratic formula:

$\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3}$

This simplifies to:

$\cos x = \frac{-5 \pm \sqrt{25 + 24}}{6} = \frac{-5 \pm \sqrt{49}}{6} = \frac{-5 \pm 7}{6}$

Calculating the two possible values:

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

Now, we find the angles for (\cos x = \frac{1}{3}) within the range (0 ≤ x < 2π):

Using a calculator, we find:

$x \approx 1.23096$ (1st quadrant) and $x \approx 5.0521$ (4th quadrant)

Rounding to three significant figures:

• First solution: $x \approx 1.23$
• Second solution: $x \approx 5.05$.

Starting with the identity:

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

This can be combined:

$= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}$

Utilizing the double angle identity, we know:

$\sin 2\theta = 2\sin \theta \cos \theta$

Thus:

$\lambda = 2$ and we can express this as:

$\tan \theta + \cot \theta = \frac{2}{\sin 2\theta} = 2 \csc 2\theta$

Consequently, the value of the constant λ is:

$λ = 2$.