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 10 - 2013 - Paper 4

We start with the given equation:

$$sin θ tan θ = 3 cos θ + 2$$

We can express tan θ as ( \frac{sin θ}{cos θ} ), leading to:

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

This can be rewritten as:

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

Multiplying through by cos θ to eliminate the denominator:

$$sin^2 θ = 3 cos^2 θ + 2 cos θ$$

Using the identity ( sin^2 θ = 1 - cos^2 θ ) gives:

$$1 - cos^2 θ = 3 cos^2 θ + 2 cos θ$$

Rearranging gives:

$$1 = 4 cos^2 θ + 2 cos θ$$

Which leads us to:

$$4 cos^2 θ + 2 cos θ - 1 = 0$$

Thus, we have proven the equation as required.

From part (a), we have the quadratic equation:

$$4 cos^2 θ + 2 cos θ - 1 = 0$$

We can solve this using the quadratic formula:

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

Here, a = 4, b = 2, and c = -1:

$$cos θ = \frac{-2 \pm \sqrt{2^2 - 4(4)(-1)}}{2(4)} = \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-2 \pm \sqrt{20}}{8}$$

This simplifies to:

$$cos θ = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}$$

Calculating the two potential solutions:

1. $$ cos θ = \frac{-1 + \sqrt{5}}{4} \ \text{(valid solution)} $$
2. $cos θ = \frac{-1 - \sqrt{5}}{4} \\ \text{(not valid, as this exceeds the range)}$

For the valid solution: Using a calculator, let’s find the approximate value: $heta ext{ corresponding to } cos^{-1}(\frac{-1 + \sqrt{5}}{4})$ gives:

• Principal solution: 72°
• From cosine’s symmetry: 144°

Thus, the solutions for 0 ≤ θ < 360° are:

• 72°
• 144°