O-LevelAdditional MathematicsTrigonometry4037/22 · May/June 2019 · Q98 Marks

Solve `6 sin²x - 13 cosx = 1` for `0° ≤ x ≤ 360°`.

(a)Solve `6 sin²x - 13 cosx = 1` for `0° ≤ x ≤ 360°`.[4]
(b)(i)Show that, for `-π/2 < y < π/2`, `(4 tan y) / (√(1+tan²y))` can be written in the form `a sin y`, where a is an integer.[3]
(b)(ii)Hence solve `(4 tan y) / (√(1+tan²y)) + 3 = 0` for `-π/2 < y < π/2` radians.[1]

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

[B1] (a) 6(1-cos²x) - 13cosx=1 oe
[M1] Solves or factorises their 3-term quadratic
[A2] 70.5 and 289.5 with no extras in range (A1 for either, ignoring extras in range)
[M1] (b)(i) Numerator: Substitution of tan y = sin y / cos y
[M1] Denominator: Substitution of 1 + tan²y = sec²y or 1 + sin²y / cos²y and correct rearrangement to 1 / cos²y oe
[A1] Correct completion to 4sin y cao
[B1] (b)(ii) -0.848[06...] rot to 3 or more figures with no extras in range

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Topic: Trigonometry, O-Level Additional Mathematics (syllabus 4037). Structured question, 8 marks.
Source: O-Level Additional Mathematics, May/June 2019 Paper 2 Variant 2, question 9.

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