(a) Show that 3cos(2θ) - sin(2θ) ≡ Rcos(2θ + α), where R > 0 and 0 < α < π/2, giving the value of R and α to 3 decimal places. (b) Hence solve, for 0 < θ < π, the equation 3cos(2θ) - sin(2θ) = 2, giving your answers to 2 decimal places.

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

1. M1: For using Rcos(2θ + α) = Rcos(2θ)cos(α) - Rsin(2θ)sin(α) and equating coefficients: Rcos(α) = 3 and Rsin(α) = 1.
2. A1: For R = √(3^2 + 1^2) = √10 (accept 3.162).
3. M1: For tan(α) = 1/3 or tan(α) = -1/-3 or tan(α) = -(-1)/3.
4. A1: For α = 0.322 radians (3 d.p.).
5. B1ft: For writing the equation as √10 cos(2θ + 0.322) = 2. (Follow through their R and α)
6. M1: For cos(2θ + 0.322) = 2/√10 and finding the principal value arccos(2/√10) = 0.887 (or 0.8868...).
7. A1: For 2θ + 0.322 = 0.887 or 2θ + 0.322 = 2π - 0.887 (or 2π + 0.887 etc.). This implies finding at least one correct general solution.
8. A1: For one correct value of θ (e.g., θ = 0.28).
9. A1: For the second correct value of θ (e.g., θ = 2.54). Both to 2 d.p. and within the range 0 < θ < π.