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 5 - 2014 - Paper 6

From the previous part, we have established the equation:

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

Next, we apply the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where:

• $a = 3$
• $b = 5$
• $c = -2$.

Calculating the discriminant: $D = b^2 - 4ac = 5^2 - 4(3)(-2) = 25 + 24 = 49.$

Now substituting into the formula gives:

$\cos x = \frac{-5 \pm \sqrt{49}}{2(3)} = \frac{-5 \pm 7}{6}.$

The possible values are:

• $\cos x = \frac{2}{6} = \frac{1}{3}$
• $\cos x = \frac{-12}{6} = -2$ (not possible since it exceeds range).

Now solving for $x$:

$x = \cos^{-1}(\frac{1}{3})$ leads to:

• $x = 1.23096$.
  This solution in the range $0 ≤ x < 2 ext{π}$ provides:
• $x = 1.23096$ and its symmetric counterpart in the second quadrant,
  which gives:
• $x = 2\pi - 1.23096 = 5.05221$.

Thus, the approximate answers to 3 significant figures are:

• $x ≈ 1.23$
• $x ≈ 5.05$.

We start with the left-hand side:
$\tan θ + \cot θ = \frac{\sin θ}{\cos θ} + \frac{\cos θ}{\sin θ}$.
This can be combined as:

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

We know from trigonometric identities that: $\sin 2θ = 2 \sin θ \cos θ$.

Thus, we rewrite the expression as: $\frac{2}{\sin 2θ}.$

Setting this equal to $λ cosec 2θ$ gives: $\tan θ + \cot θ = \frac{2}{\sin 2θ} = λ cosec 2θ.$
So we identify that $λ = 2.$
This holds true for all θ not equal to $n\frac{\pi}{2}$, ensuring we do not involve division by zero.