In this question solutions based entirely on graphical or numerical methods are not acceptable - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 4

First, let's rearrange the equation:

$6 \cos^2 \theta - 6 \sin \theta - 5 - \sin \theta = 0$

Using the Pythagorean identity, substitute $\cos^2 \theta$:

$6(1 - \sin^2 \theta) - 6\sin \theta - 5 - \sin \theta = 0$

This simplifies to:

$6 - 6\sin^2 \theta - 7\sin \theta - 5 = 0$

Rearranging gives:

$6\sin^2 \theta + 7\sin \theta - 1 = 0$

Solve this quadratic equation using the quadratic formula:

$\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 6$, $b = 7$, $c = -1$:

$\sin \theta = \frac{-7 \pm \sqrt{7^2 - 4\cdot6\cdot(-1)}}{2\cdot6} = \frac{-7 \pm \sqrt{49 + 24}}{12} = \frac{-7 \pm \sqrt{73}}{12}$

Calculating the values gives:

$\sin \theta \approx 0.253, -0.598$

For each value, derive $\theta$:

For $\sin \theta = 0.253$:

$\theta \\approx \arcsin(0.253) = 0.256\ ext{ radians}$

The second solution in range:

$\theta \\approx \pi - 0.256 = 2.885\ ext{ radians}$

For $\sin \theta = -0.598$:

$\theta \\approx \arcsin(-0.598) = -0.643$

Adjusting this to the range gives:

$2\pi + (-0.643) = 5.640, 3.785\ (2\pi - 0.643)$

The solutions in radians to 3 significant figures are:

$\theta \\approx 0.256, 2.885, 3.785, 5.640$