8. (i) Solve, for -180° ≤ x < 180°, tan(x - 40°) = 1.5 giving your answers to 1 decimal place - Edexcel - A-Level Maths Pure - Question 1 - 2012 - Paper 4

Starting with the equation:

$ext{sin}θ \, ext{tan}θ = 3\text{cos}θ + 2$

We can substitute tanθ with sinθ/cosθ:

$ext{sin}θ \cdot \frac{\text{sin}θ}{\text{cos}θ} = 3\text{cos}θ + 2$

This simplifies to:

$\frac{\text{sin}^2 θ}{\text{cos}θ} = 3\text{cos}θ + 2$

Multiplying throughout by cos²θ gives us:

$(\text{sin}^2 θ = 3\text{cos}^2 θ + 2\text{cos}θ)$

Using the identity $\text{sin}^2 θ = 1 - \text{cos}^2 θ$:

$1 - \text{cos}^2 θ = 3\text{cos}^2 θ + 2\text{cos}θ$

Rearranging yields:

$4\text{cos}^2 θ + 2\text{cos}θ - 1 = 0$

From the previous part, we have a quadratic equation in terms of cosθ:

$4\text{cos}^2 θ + 2\text{cos}θ - 1 = 0$

Using the quadratic formula:

$\text{cos}θ = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where a = 4, b = 2, c = -1:

$\text{cos}θ = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4}$

$= \frac{-2 \pm \sqrt{4 + 16}}{8}$

$= \frac{-2 \pm \sqrt{20}}{8}$

$= \frac{-2 \pm 2\sqrt{5}}{8}$

$= \frac{-1 \pm \sqrt{5}}{4}$

Calculating the values:

1. $\text{cos}θ = \frac{-1 + \sqrt{5}}{4} \approx 0.618$ (Choose this since cosθ must be between -1 and 1)
2. $\text{cos}θ = \frac{-1 - \sqrt{5}}{4} \approx -0.866$

Now, finding the angles:

For $\text{cos}θ ≈ 0.618$:
$θ ≈ 72°$ For $\text{cos}θ ≈ -0.866$:
$θ ≈ 144°$

The final solutions are: $θ = 72° \, \text{and} \, 144°$