| Exam Board | SPS |
|---|---|
| Module | SPS FM (SPS FM) |
| Year | 2026 |
| Session | November |
| Marks | 8 |
| Topic | Standard trigonometric equations |
| Type | Deduce related solution |
| Difficulty | Standard +0.3 This is a straightforward trigonometric equation problem requiring standard algebraic manipulation (converting tan to sin/cos, using Pythagorean identity) followed by solving a quadratic in cos θ and applying a compound angle substitution. While it has multiple steps and requires careful algebra, all techniques are routine A-level methods with no novel insight needed. The 8 marks reflect length rather than conceptual difficulty, placing it slightly above average. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
\begin{enumerate}[label=(\alph*)]
\item Show that the equation
$$4\cos\theta - 1 = 2\sin\theta\tan\theta$$
can be written in the form
$$6\cos^2\theta - \cos\theta - 2 = 0$$
[4]
\item Hence solve, for $0 \leq x < 90°$
$$4\cos 3x - 1 = 2\sin 3x\tan 3x$$
giving your answers, where appropriate, to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.)
[4]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM 2026 Q5 [8]}}