Show that the equation cot² θ + 2 cos 2θ = 4 can be written in the form 4 sin⁴ θ + 3 sin² θ − 1 = 0.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

1. [M1] (a) Use correct Pythagoras cot²θ = cosec²θ–1 or cot²θ = 1/sin²θ–1 or cot²θ= cos²θ/sin²θ and then cos²θ= 1 − sin²θ, together with double angle formula cos2θ = 1 – 2sin²θ, to obtain an equation in sin²θ or sin θ and cosec²θ. If consistent omission of brackets, e.g. (sinθ)² written as sinθ² then SC B1 in place of M1A1.
2. [A1] (a) Obtain a correct equation in sin θ in any form, e.g. 1/sin²θ - 1 + 2(1 – 2sin²θ)=4 or (1-sin²θ)/sin²θ +2(1-2sin²θ)=4. If cos²θ/sin²θ +2(1-2sin²θ) = 4 then e.g. 1-sin²θ+2(1-2sin²θ)sin²θ = 4. (missing sin² on right) allow M1A1A0.
3. [A1] (a) Reduce to the given answer of 4sin⁴ θ+3sin² θ-1=0 correctly. AG Must follow from a horizontal equation (no denominators). If s = sin θ used and defined, allow all marks. If not defined, award M1A1A0.
4. [M1] (b) Solve the given quadratic to obtain a value for θ. (4sin²θ - 1)(sin²θ + 1) = 0 and solve for θ. Incorrect sign in solution of quadratic seen, e.g. (4sin²θ - 1)(sin²θ – 1) = 0 then M0 A0 A0 but if only see (4sin²θ- 1) = 0 and nothing incorrect seen allow 3/3.
5. [A1] (b) Obtain answer, e.g. θ = 30°. π/6 award A0.
6. [A1] (b) Obtain three further answers, e.g. θ = 150°, 210° and 330° and no others in the interval. Ignore any answers outside interval. 5π/6 7π/6 11π/6 award A1.