A-Level Mathematics October/November 2024 Q4: Solve the equation 4 sin⁴θ + 12 sin²θ – 7 = 0 for 0° ≤ θ ≤ 360°.

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  The given equation is $4 \sin^4\theta + 12 \sin^2\theta – 7 = 0$ for $0° \le \theta \le 360°$.

  This equation is a quadratic in the form of $\sin^2\theta$. Let $x = \sin^2\theta$. The equation becomes: $4x^2 + 12x - 7 = 0$

  We can solve for $x$ using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-12 \pm \sqrt{12^2 - 4(4)(-7)}}{2(4)}$ $x = \frac{-12 \pm \sqrt{144 + 112}}{8}$ $x = \frac{-12 \pm \sqrt{256}}{8}$ $x = \frac{-12 \pm 16}{8}$

  This gives two possible values for $x$: $x = \frac{-12 + 16}{8} = \frac{4}{8} = \frac{1}{2}$ $x = \frac{-12 - 16}{8} = \frac{-28}{8} = -\frac{7}{2}$

  Now, we substitute back $\sin^2\theta$ for $x$. Case 1: $\sin^2\theta = \frac{1}{2}$ Case 2: $\sin^2\theta = -\frac{7}{2}$

  Case 2 is not possible, as $\sin^2\theta$ must be non-negative. We reject this solution.

  From Case 1, we have $\sin^2\theta = \frac{1}{2}$. Taking the square root of both sides gives: $\sin\theta = \pm\sqrt{\frac{1}{2}} = \pm\frac{1}{\sqrt{2}}$

  The basic angle, $\alpha$, is $\sin^{-1}(\frac{1}{\sqrt{2}}) = 45°$.

  We need to find solutions for $\sin\theta = \frac{1}{\sqrt{2}}$ and $\sin\theta = -\frac{1}{\sqrt{2}}$ in the range $0° \le \theta \le 360°$.

  For $\sin\theta = \frac{1}{\sqrt{2}}$ (positive, Quadrants 1 and 2): $\theta = \alpha = 45°$ $\theta = 180° - \alpha = 180° - 45° = 135°$

  For $\sin\theta = -\frac{1}{\sqrt{2}}$ (negative, Quadrants 3 and 4): $\theta = 180° + \alpha = 180° + 45° = 225°$ $\theta = 360° - \alpha = 360° - 45° = 315°$

  Final solutions are $\theta = 45°, 135°, 225°, 315°$.

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  How the marks are awarded

  • M1 — Awarded for recognising the equation as a quadratic in $\sin^2\theta$ and attempting to solve it, for instance by using the quadratic formula on $4x^2 + 12x - 7 = 0$.
  • M1 — Awarded for correctly finding $\sin^2\theta = 1/2$ (and rejecting the other solution), and then taking the square root to find $\sin\theta = \pm 1/\sqrt{2}$.
  • A1 — Awarded for finding any two of the correct final angles. For example, finding $\theta = 45°$ and $\theta = 135°$ from $\sin\theta = 1/\sqrt{2}$.
  • A1 — Awarded for finding all four correct angles ($45°, 135°, 225°, 315°$) and no other solutions within the specified range.

  Common mistakes

  • Forgetting the negative root when solving $\sin^2\theta = 1/2$, leading to $\sin\theta = 1/\sqrt{2}$ only and finding just two solutions.
  • Making a calculation error in the quadratic formula, such as with the discriminant ($b^2 - 4ac$), which prevents finding the correct value for $\sin^2\theta$.
  • Finding the correct basic angle (45°) but then making errors with the quadrant rules, for example calculating $360° + 45°$ or $270° - 45°$.
  • Failing to reject the solution $\sin^2\theta = -7/2$ and attempting to take its square root, leading to confusion and incorrect working.

  Examiner tip: Practice identifying and solving equations in 'disguised' quadratic form by making a suitable substitution for the trigonometric function.

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