9. (a) Prove that sin 2x - tan x = tan x cos 2x, x ≠ (2n + 1)90°, n ∈ Z (4) (b) Given that x ≠ 90° and x ≠ 270°, solve, for 0 ≤ x < 360°, sin 2x - tan x = 3 tan x sin x - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 4

Starting from the equation:

$$sin 2x - tan x = 3 tan x sin x$$

Substituting (sin 2x = 2sin x cos x) gives:

$$2sin x cos x - \frac{sin x}{cos x} = 3 \frac{sin x}{cos x} sin x$$

Multiplying through by (cos x) to eliminate the denominator:

$$2sin x cos^2 x - sin x = 3sin^2 x$$

Rearranging yields:

$$2sin x cos^2 x - 3sin^2 x - sin x = 0$$

Factoring out sin x:

$$sin x(2cos^2 x - 3sin x - 1) = 0$$

Therefore, (sin x = 0) gives solutions at:

• (x = 0°, 180°)

For the quadratic equation in terms of (sin x), using the quadratic formula:

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

Where:

• (a = -3, b = -1, c = 2)

Calculating the discriminant (b^2 - 4ac):

$$(-1)^2 - 4(-3)(2) = 1 + 24 = 25$$

Thus:

$$sin x = \frac{1 \pm 5}{-6}$$

This gives:

• (sin x = -\frac{2}{3}) (not valid in the given range)
• (sin x = \frac{1}{3})

Then, using (sin^{-1}(\frac{1}{3})) gives:

• Approximate values of (x = 19.1°) and its supplementary angle in the range (0 ≤ x < 360°) is:
• (x = 180° - 19.1° = 160.9°)

Thus, the solutions are:

• (x = 0°, 180°, 19.1°, 160.9°)