In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 11 - 2020 - Paper 2

Starting with the equation:

$$1 - ext{cos} 3x = ext{sin} x$$

From part (a) we know:

$$ext{cos} 3x = 4 ext{cos}^2 x - 3 ext{cos} x$$

Substituting gives us:

$$1 - (4 ext{cos}^2 x - 3 ext{cos} x) = ext{sin} x$$

This simplifies to:

$$1 - 4 ext{cos}^2 x + 3 ext{cos} x = ext{sin} x$$

Rearranging leads to:

$$4 ext{cos}^2 x - 3 ext{cos} x + ( ext{sin} x - 1) = 0$$

Applying the identity $ext{sin}^2 x + ext{cos}^2 x = 1$ helps to rewrite:

$$= 4(1 - ext{sin}^2 x) - 3 ext{cos} x + ( ext{sin} x - 1) = 0$$

We will need to solve this equation over the interval $-90^{ extcirc} ext{ to } 180^{ extcirc}$. Using appropriate approaches, we find the solutions for $x = -90°, 0°, 90°, 139°$.