A-Level Mathematics May/June 2025 Q5: Solve the equation 4 sin θ tan θ = 1 + 5 cos θ for -180° < θ < 180°.

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  The equation to be solved is $4 \sin \theta \tan \theta = 1 + 5 \cos \theta$ for $-180° < \theta < 180°$.

  First, express the equation in terms of $\sin \theta$ and $\cos \theta$ by using the identity $\tan \theta \equiv \frac{\sin \theta}{\cos \theta}$. $4 \sin \theta \left( \frac{\sin \theta}{\cos \theta} \right) = 1 + 5 \cos \theta$

  rac{4 \sin^2 \theta}{\cos \theta} = 1 + 5 \cos \theta

  Multiply by $\cos \theta$ to clear the fraction (assuming $\cos \theta \neq 0$): $4 \sin^2 \theta = \cos \theta (1 + 5 \cos \theta)$ $4 \sin^2 \theta = \cos \theta + 5 \cos^2 \theta$

  Now, use the identity $\sin^2 \theta \equiv 1 - \cos^2 \theta$ to form an equation solely in terms of $\cos \theta$. $4(1 - \cos^2 \theta) = \cos \theta + 5 \cos^2 \theta$

  Expand and rearrange to form a quadratic equation in $\cos \theta$. $4 - 4 \cos^2 \theta = \cos \theta + 5 \cos^2 \theta$ $0 = 9 \cos^2 \theta + \cos \theta - 4$

  Let $c = \cos \theta$. The equation is $9c^2 + c - 4 = 0$. We solve this using the quadratic formula, $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. $\cos \theta = \frac{-1 \pm \sqrt{1^2 - 4(9)(-4)}}{2(9)}$ $\cos \theta = \frac{-1 \pm \sqrt{1 + 144}}{18}$ $\cos \theta = \frac{-1 \pm \sqrt{145}}{18}$

  This gives two possible values for $\cos \theta$: $\cos \theta = \frac{-1 + \sqrt{145}}{18} = 0.61343...$ $\cos \theta = \frac{-1 - \sqrt{145}}{18} = -0.72454...$

  For each value, we find the corresponding angles for $\theta$ in the range $-180° < \theta < 180°$.

  Case 1: $\cos \theta = 0.61343...$ The principal value is $\theta = \cos^{-1}(0.61343...) = 52.16°$. Since $\cos \theta$ is positive, solutions are in the first and fourth quadrants. In the given range, this corresponds to $\theta = 52.2°$ and $\theta = -52.2°$ (to 1 d.p.).

  Case 2: $\cos \theta = -0.72454...$ The principal value is $\theta = \cos^{-1}(-0.72454...) = 136.42°$. Since $\cos \theta$ is negative, solutions are in the second and third quadrants. In the given range, this corresponds to $\theta = 136.4°$ and $\theta = -136.4°$ (to 1 d.p.).

  The solutions in the required range are $\theta = -136.4°, -52.2°, 52.2°, 136.4°$.

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  How the marks are awarded

  • M1 — Correctly replacing tan θ with sin θ / cos θ in the first step of the working.
  • M1 — Using the identity sin²θ = 1 - cos²θ and successfully rearranging the equation to form a three-term quadratic in cos θ.
  • A1 — Obtaining the correct quadratic equation 9 cos²θ + cos θ - 4 = 0.
  • DM1 — Applying a valid method (e.g., the quadratic formula) to solve their three-term quadratic for cos θ, leading to two numerical values.
  • A1 — Finding any two of the four correct final answers for θ, correctly rounded. For example, finding 52.2° and 136.4°.
  • A1 — Stating all four correct solutions (±52.2°, ±136.4°) and no other incorrect solutions within the specified range of -180° < θ < 180°.

  Common mistakes

  • Making a sign error when rearranging the equation to form the quadratic, for example getting 9cos²θ + cosθ + 4 = 0, which has no real solutions for cos θ.
  • Incorrectly finding the second angle for a given value of cos θ. For example, for cos θ = 0.6134..., finding θ = 52.2° and then calculating 180° - 52.2° instead of -52.2° for the given range.
  • Finding only the two principal values from the calculator (52.2° and 136.4°) and failing to find the other two solutions within the negative part of the range.
  • Rounding prematurely, for example rounding the values of cos θ to two decimal places before finding the angle θ, which can lead to final answers outside the acceptable accuracy range.

  Examiner tip: When solving trigonometric equations with multiple different trig functions, aim to use identities to express the entire equation in terms of a single function, which often leads to a solvable quadratic.

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