A-Level Mathematics October/November 2024 Q8(b): Solve the equation sin²θ+2cos²θ = 4 sinθ+3 for 0° < θ < 360°.

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  The given equation is: $sin^2\theta + 2cos^2\theta = 4sin\theta + 3$

  To solve this, we first express the equation in terms of a single trigonometric function. We use the identity $cos^2\theta = 1 - sin^2\theta$.

  $sin^2\theta + 2(1 - sin^2\theta) = 4sin\theta + 3$

  Expand the brackets: $sin^2\theta + 2 - 2sin^2\theta = 4sin\theta + 3$

  Simplify and collect terms to form a quadratic equation in $sin\theta$: $-sin^2\theta + 2 = 4sin\theta + 3$

  Rearrange all terms to one side: $0 = sin^2\theta + 4sin\theta + 3 - 2$ $sin^2\theta + 4sin\theta + 1 = 0$

  This is a quadratic equation that does not factorise easily, so we use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, with $x = sin\theta$, $a=1$, $b=4$, and $c=1$.

  $sin\theta = \frac{-4 \pm \sqrt{4^2 - 4(1)(1)}}{2(1)}$ $sin\theta = \frac{-4 \pm \sqrt{16 - 4}}{2}$ $sin\theta = \frac{-4 \pm \sqrt{12}}{2}$ $sin\theta = \frac{-4 \pm 2\sqrt{3}}{2}$ $sin\theta = -2 \pm \sqrt{3}$

  This gives two possible values for $sin\theta$:

  1. $sin\theta = -2 + \sqrt{3} \approx -0.2679$
  2. $sin\theta = -2 - \sqrt{3} \approx -3.732$

  Since the range of $sin\theta$ is $[-1, 1]$, the value $sin\theta \approx -3.732$ has no solutions.

  We solve for $sin\theta = -0.2679...$ First, find the principal value (basic angle): $\alpha = sin^{-1}(0.2679...) \approx 15.5^\circ$

  Since $sin\theta$ is negative, the solutions for $\theta$ lie in the third and fourth quadrants for the range $0^\circ < \theta < 360^\circ$.

  Third quadrant solution: $\theta = 180^\circ + \alpha = 180^\circ + 15.5^\circ = 195.5^\circ$

  Fourth quadrant solution: $\theta = 360^\circ - \alpha = 360^\circ - 15.5^\circ = 344.5^\circ$

  Both solutions are within the required range.

  Final answers: $\theta = 195.5^\circ, 344.5^\circ$

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  How the marks are awarded

  • *M1 — For using the correct identity $cos^2\theta = 1 - sin^2\theta$ to transform the equation into one containing only the $sin\theta$ term, leading to a three-term quadratic.
  • A1 — For correctly simplifying and rearranging the equation to obtain the quadratic $sin^2\theta + 4sin\theta + 1 = 0$.
  • DM1 — For making a valid attempt to solve their three-term quadratic equation in $sin\theta$, typically by using the quadratic formula. This mark is dependent on the first M1 mark being awarded.
  • A1 — For obtaining the first correct solution, $\theta = 195.5^\circ$. This requires a correct method from a correct value of $sin\theta$.
  • A1 FT — For obtaining the second correct solution, $\theta = 344.5^\circ$. This mark is awarded for correctly finding the second angle in the correct quadrant, and can be a follow-through from a correct method on a slightly inaccurate principal value.

  Common mistakes

  • Using an incorrect identity, such as $cos^2\theta = 1 + sin^2\theta$, which leads to an incorrect quadratic and the loss of all subsequent accuracy marks.
  • Making an algebraic sign error when rearranging the equation, for example obtaining $sin^2\theta - 4sin\theta - 1 = 0$, which leads to incorrect values for $sin\theta$ and the final angles.
  • Finding only one solution after calculating $sin\theta$, or finding the second solution in an incorrect quadrant (e.g., $180^\circ - \alpha$ instead of $180^\circ + \alpha$).
  • Having the calculator in radians mode, leading to incorrect angle values (e.g., 3.41 and 5.99) instead of degrees.

  Examiner tip: When an equation contains both $sin^2\theta$ and $cos\theta$ (or $cos^2\theta$ and $sin\theta$), always use the identity $sin^2\theta + cos^2\theta = 1$ to convert it into a quadratic equation in a single trigonometric function.

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