Maths with David

    9709. P1. Trigonometry. Past Exam Questions

    November 2025 (9709/12). Question 6

    (a) Sketch the graph of y = 3 sin x + 2 for 0 \leq x \leq 2 \pi [2 marks]

    (b) Determine the number of solutions in the interval 0 \leq x \leq 2 \pi of each of the following equations:

    • 3 sin x + 2 = x [1 mark]
    • 3 sin x + 2 = 5 – x [1 mark]

    (c) Solve the equation 3 sin x + 2 = 5 cos2x – 1 for 0 \leq x \leq 2 \pi $ [5 marks]

    9709/11/M/J/25q1 – Mark Scheme

    Solve the equation 6sin \theta = 1 + \frac{2}{ sin \theta} for -180^{ \circ } < \theta < 180^{ \circ} [4 marks]

    9709/12/M/J/25q5

    The equation of a curve is y = 4cos2x + 3 for 0 \leq x \leq 2 \pi

    (a) State the greatest and least possible values of y. [2 marks]

    (b.) Sketch the curve. [2 marks]

    (c) Hence determine the number of solutions of the equation 4cos2x + 3 = 2x – 1 for 0 \leq x \leq 2 \pi . [1 mark]

    9709/12/M/J/25q7

    (a) Prove the identity \frac{ tan \theta + 7 }{ tan^2 \theta - 3 } \equiv \frac{ sin \theta cos \theta + 7 cos^2 \theta }{ 1 - 4 cos^2 \theta } [3 marks]

    (b) Hence solve the equation \frac{ sin \theta cos \theta + 7 cos^2 \theta}{ 1 - 4cos^2 \theta } = \frac{5}{ tan \theta} for 0^{ \circ} \leq \theta \leq 180^{ \circ}. [4 marks]

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