The variables x and θ satisfy the differential equation dx/dθ sin 2θ = (4x+3) cos 2θ, and x = 0 when θ = ¹⁄₁₂π. Solve the differential equation and obtain an expression for x in terms of θ.

Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

1. [B1] Separate variables correctly ∫1/(4x+3) dx = ∫cos2θ/sin2θ dθ Can be implied by obtaining both correct integrals.
2. [B1] Obtain term 1/4 ln(4x+3) OE or ln(x+3/4).
3. [M1] Obtain term of the form A ln(sin2θ) Or A ln (k sin 2θ) OE, e.g. P ln a sin θ+ Q ln b cos θ from using the tan 2θ formula. Or expanding cos 2θ as cos² θ – sin² θ.
4. [A1] Obtain term 1/2 ln(sin2θ) OE Correct in any form, e.g. 1/2 ln a sin θ + 1/2 ln b cos θ.
5. [M1] Use x = 0 when θ = 1/12π to evaluate a constant or as limits in a solution containing terms of the form ln(sin 2θ) and ln (4x + 3) c = ... seen or implied E.g. c=ln3−1/2lnsin(π/6) ln(4x+3)−ln3= 1/2ln(sin 2θ)−1/2ln(1/2). Note that the constant may be expressed as a logarithm
6. [A1] Obtain correct answer in any form with the trigonometry evaluated E.g. ln(4x+3) = 1/2ln(sin2θ) + 1/2ln12 ln(4x+3)=ln(sin2θ)+0.621 1/4 ln(4x+3)=ln(2sin2θ)
7. [A1] Obtain final answer x = (12sin^2 2θ-3)/4 OE Allow (e^(2.48....)sin^2 2θ-3)/4. Allow 2.48. Allow x = 1/4 (e^(2ln(sin2θ)+2.48...) - 3).