In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 10 - 2020 - Paper 2

Starting with the equation:

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

We utilize the identity for $ext{cos } 3x$:

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

Substituting this into the equation gives us:

$$1 - (4 ext{cos}^3 x - 3 ext{cos} x) = ext{sin } x \\ \\ ext{or, equivalently,} \\ \ 1 - 4 ext{cos}^3 x + 3 ext{cos} x = ext{sin } x$$

Rearranging this yields a cubic equation:

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

At this point, substituting the known limits for $x$:

1. For $x = -90^{ ext{o}}$, we see that:
   • LHS = 1, RHS = 0 ( not a solution).
2. For $x = 0^{ ext{o}}$:
   • LHS = 1, RHS = 0 (not a solution).
3. For $x = 90^{ ext{o}}$:
   • LHS = 1, RHS = 0 (not a solution).
4. Testing within the range:
   • If $x = 180^{ ext{o}}$ then we get:
   • LHS = -1, RHS = 0 (not a solution).

Thus providing the necessary roots based on the cubic relation within the specified limits.