A-Level Mathematics May/June 2025 Q4: Solve the equation 3 cotx-4 cot2x = 3 for 0° ≤ x ≤ 180°.

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  accepted ✓

  The given equation is $3 \cot x - 4 \cot 2x = 3$ for $0° \le x \le 180°$.

  First, we express the equation in terms of $\tan x$. We use the identities $\cot x = \frac{1}{\tan x}$ and $\cot 2x = \frac{1}{\tan 2x} = \frac{1 - \tan^2 x}{2 \tan x}$.

  Substituting these into the equation gives: $3\left(\frac{1}{\tan x}\right) - 4\left(\frac{1 - \tan^2 x}{2 \tan x}\right) = 3$

  Simplify the second term: $\frac{3}{\tan x} - \frac{2(1 - \tan^2 x)}{\tan x} = 3$

  To clear the denominator, multiply the entire equation by $\tan x$ (noting that $\tan x \neq 0$): $3 - 2(1 - \tan^2 x) = 3 \tan x$

  Expand the brackets: $3 - 2 + 2 \tan^2 x = 3 \tan x$ $1 + 2 \tan^2 x = 3 \tan x$

  Rearrange this into a standard quadratic equation in the form $ax^2+bx+c=0$, where the variable is $\tan x$: $2 \tan^2 x - 3 \tan x + 1 = 0$

  This is a quadratic equation that can be factorised: $(2 \tan x - 1)(\tan x - 1) = 0$

  This gives two possible solutions for $\tan x$: $2 \tan x - 1 = 0 \implies \tan x = \frac{1}{2}$ OR $\tan x - 1 = 0 \implies \tan x = 1$

  Now we solve for $x$ in the given interval $0° \le x \le 180°$.

  Case 1: $\tan x = 1$ $x = \tan^{-1}(1) = 45°$ This is within the required range.

  Case 2: $\tan x = \frac{1}{2}$ $x = \tan^{-1}(0.5) = 26.565...°$ Rounding to one decimal place, $x = 26.6°$. This is also within the required range.

  The solutions are $x = 26.6°$ and $x = 45°$.

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  How the marks are awarded

  • M1 — Using the correct identities for cot(x) and cot(2x) to write the equation entirely in terms of tan(x), as shown in the step: $3(\frac{1}{\tan x}) - 4(\frac{1 - \tan^2 x}{2 \tan x}) = 3$.
  • A1 — Obtaining a correct equation with the denominator cleared, such as $3 - 2(1 - \tan^2 x) = 3 \tan x$.
  • A1 — Correctly simplifying and rearranging the equation into the 3-term quadratic form: $2 \tan^2 x - 3 \tan x + 1 = 0$.
  • DM1 — Solving the 3-term quadratic in tan(x) to find values for tan(x) (i.e., 1 and 1/2) and then using the inverse tangent function to find at least one value for x.
  • A1 — Obtaining the exact answer $x = 45°$ from solving $\tan x = 1$.
  • A1 — Obtaining the second answer $x = 26.6°$ (or AWRT 26.6) and ensuring no other incorrect solutions are given within the specified range.

  Common mistakes

  • Using an incorrect double angle formula for tan(2x), such as $\frac{2\tan x}{1+\tan^2 x}$, which leads to an incorrect quadratic equation.
  • Making an algebraic error when simplifying, for example expanding $-2(1 - \tan^2 x)$ to $-2 - 2\tan^2 x$ instead of $-2 + 2\tan^2 x$.
  • Incorrectly factorising the quadratic or making an error in the quadratic formula, leading to wrong values for tan(x).
  • Finding additional incorrect solutions outside the given range of $0° \le x \le 180°$, for example by adding 180° to the principal values.

  Examiner tip: Master expressing all trigonometric terms using a single function and angle (like tan x) to reduce the equation to a solvable polynomial.

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