SPS SPS SM 2022 January — Question 8 8 marks

Exam BoardSPS
ModuleSPS SM (SPS SM)
Year2022
SessionJanuary
Marks8
TopicTrigonometric equations in context
TypeShow then solve substituted equation
DifficultyStandard +0.3 This is a straightforward two-part question requiring standard algebraic manipulation (converting tan to sin/cos, using sin²+cos²=1) followed by solving a quadratic in cos θ and applying the result to a compound angle. The techniques are routine for A-level and the question provides significant scaffolding through part (a).
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals

8.
  1. Show that the equation $$4 \cos \theta - 1 = 2 \sin \theta \tan \theta$$ can be written in the form $$6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0$$
  2. Hence solve, for \(0 \leqslant x < 90 ^ { \circ }\) $$4 \cos 3 x - 1 = 2 \sin 3 x \tan 3 x$$ giving your answers, where appropriate, to one decimal place.
    (Solutions based entirely on graphical or numerical methods are not acceptable.)
    [0pt]

8.\\
\begin{enumerate}[label=(\alph*)]
\item Show that the equation

$$4 \cos \theta - 1 = 2 \sin \theta \tan \theta$$

can be written in the form

$$6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0$$
\item Hence solve, for $0 \leqslant x < 90 ^ { \circ }$

$$4 \cos 3 x - 1 = 2 \sin 3 x \tan 3 x$$

giving your answers, where appropriate, to one decimal place.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\[0pt]
\\
\end{enumerate}

\hfill \mbox{\textit{SPS SPS SM 2022 Q8 [8]}}