SPS SPS SM Pure 2024 June — Question 11 7 marks

Exam BoardSPS
ModuleSPS SM Pure (SPS SM Pure)
Year2024
SessionJune
Marks7
TopicTrigonometric equations in context
TypeShow then solve substituted equation
DifficultyStandard +0.3 This is a standard A-level trig equation question with clear scaffolding. Part (a) requires routine algebraic manipulation (expanding, using tan = sin/cos, and the Pythagorean identity), while part (b) applies the given result with a straightforward substitution θ = 2x. The quadratic formula and range restriction are standard techniques. Slightly easier than average due to the helpful structure.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals

  1. In this question you must show detailed reasoning.
\section*{Solutions relying entirely on calculator technology are not acceptable.}
  1. Show that the equation $$( 3 \cos \theta - \tan \theta ) \cos \theta = 2$$ can be written as $$3 \sin ^ { 2 } \theta + \sin \theta - 1 = 0$$
  2. Hence solve for \(- \frac { \pi } { 2 } \leq x \leq \frac { \pi } { 2 }\) $$( 3 \cos 2 x - \tan 2 x ) \cos 2 x = 2$$

\begin{enumerate}
  \item In this question you must show detailed reasoning.
\end{enumerate}

\section*{Solutions relying entirely on calculator technology are not acceptable.}
(a) Show that the equation

$$( 3 \cos \theta - \tan \theta ) \cos \theta = 2$$

can be written as

$$3 \sin ^ { 2 } \theta + \sin \theta - 1 = 0$$

(b) Hence solve for $- \frac { \pi } { 2 } \leq x \leq \frac { \pi } { 2 }$

$$( 3 \cos 2 x - \tan 2 x ) \cos 2 x = 2$$

\hfill \mbox{\textit{SPS SPS SM Pure 2024 Q11 [7]}}