A-Level Mathematics May/June 2024 Q7(b): Solve the equation tan²θ+7 sinθ cosec 2θ = 8 for -π < θ < π.

1 answer

• ▲0
  accepted ✓

  The equation to solve is $\tan^2\theta + 7 \sin\theta \csc 2\theta = 8$ for $-\pi < \theta < \pi$.

  First, we express the equation in terms of a single trigonometric function. We use the identities $\tan^2\theta = \sec^2\theta - 1$ and $\csc 2\theta = \frac{1}{\sin 2\theta} = \frac{1}{2\sin\theta\cos\theta}$.

  The second term becomes: $7 \sin\theta \csc 2\theta = 7 \sin\theta \left( \frac{1}{2\sin\theta\cos\theta} \right) = \frac{7}{2\cos\theta} = \frac{7}{2}\sec\theta$.

  Substitute these into the original equation: $(\sec^2\theta - 1) + \frac{7}{2}\sec\theta = 8$

  Now, rearrange to form a quadratic equation in $\sec\theta$: $\sec^2\theta + \frac{7}{2}\sec\theta - 9 = 0$

  To eliminate the fraction, multiply the entire equation by 2: $2\sec^2\theta + 7\sec\theta - 18 = 0$

  This is a quadratic equation in the form $ax^2+bx+c=0$, with $x = \sec\theta$. We can solve for $\sec\theta$ using the quadratic formula: $\sec\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\sec\theta = \frac{-7 \pm \sqrt{7^2 - 4(2)(-18)}}{2(2)}$ $\sec\theta = \frac{-7 \pm \sqrt{49 + 144}}{4}$ $\sec\theta = \frac{-7 \pm \sqrt{193}}{4}$

  This gives two possible values for $\sec\theta$: $\sec\theta = \frac{-7 + \sqrt{193}}{4} \approx 1.7233$ or $\sec\theta = \frac{-7 - \sqrt{193}}{4} \approx -5.2233$

  Now we find the corresponding values for $\cos\theta = \frac{1}{\sec\theta}$: $\cos\theta = \frac{4}{-7 + \sqrt{193}} \approx 0.58028$ or $\cos\theta = \frac{4}{-7 - \sqrt{193}} \approx -0.19145$

  For each value of $\cos\theta$, we find the solutions for $\theta$ in the interval $-\pi < \theta < \pi$.

  Case 1: $\cos\theta \approx 0.58028$ The principal value is $\alpha = \arccos(0.58028) \approx 0.9515$ radians. Since $\cos\theta = \cos(-\theta)$, the other solution in the range is $-\alpha$. So, $\theta \approx 0.952$ and $\theta \approx -0.952$ (to 3 s.f.).

  Case 2: $\cos\theta \approx -0.19145$ The principal value is $\alpha = \arccos(-0.19145) \approx 1.7631$ radians. The other solution in the range is $-\alpha$. So, $\theta \approx 1.76$ and $\theta \approx -1.76$ (to 3 s.f.).

  The four solutions in the given interval are $\theta = \pm 0.952, \pm 1.76$.

  ---

  How the marks are awarded

  • *M1 — Using the identities for tan²θ and cosec 2θ to begin converting the equation into one involving only secθ or cosθ.
  • A1 — Obtaining the correct quadratic equation 2sec²θ + 7secθ - 18 = 0, or an equivalent form such as sec²θ + (7/2)secθ - 9 = 0.
  • DM1 — Applying a correct method (e.g., quadratic formula) to solve their 3-term quadratic for secθ or cosθ, and then using arccos to find at least one value for θ. This mark is dependent on the first M1 mark.
  • A1 — Obtaining any two of the four correct solutions, for example, θ = 0.952 and θ = -0.952, correctly rounded to 3 significant figures.
  • A1 — Obtaining the remaining two correct solutions, θ = 1.76 and θ = -1.76, for a complete set of four correct answers.

  Common mistakes

  • Incorrectly simplifying the term 7 sinθ cosec 2θ, for example to 7/2 or 7 secθ, by misremembering the double angle formula for sine.
  • Making an algebraic error when rearranging the equation into a quadratic, such as adding 8 instead of subtracting it.
  • Forgetting to find all four solutions. A common error is to find the principal value from the calculator and forget the corresponding negative angle (e.g., finding θ = 0.952 but not θ = -0.952).
  • Having the calculator in degrees mode instead of radians, which is required by the interval -π < θ < π.

  Examiner tip: This question rewards the systematic application of trigonometric identities to simplify a complex equation into a standard solvable form, like a quadratic.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.