WJEC Unit 3 2024 June — Question 2 11 marks

Exam BoardWJEC
ModuleUnit 3 (Unit 3)
Year2024
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeMultiple independent equations — all direct solve
DifficultyStandard +0.3 Part (a) requires using the identity cosec²θ = 1 + cot²θ to form a quadratic in cotθ, then solving for θ in the given range - a standard technique. Part (b) is a routine R-formula question requiring finding R and α, then solving a simple cosine equation. Both parts are textbook exercises with well-practiced methods, slightly above average only due to the multi-step nature and 11 total marks.
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^21.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc

  1. Find all values of \(\theta\) in the range \(0° < \theta < 360°\) satisfying $$3\cot\theta + 4\cosec^2\theta = 5.$$ [5]
  2. By writing \(24\cos x - 7\sin x\) in the form \(R\cos(x + \alpha)\), where \(R\) and \(\alpha\) are constants with \(R > 0\) and \(0° < \alpha < 90°\), solve the equation $$24\cos x - 7\sin x = 16$$ for values of \(x\) between \(0°\) and \(360°\). [6]

Question 2:
AnswerMarks
211
Question 2:
2 | 11
\begin{enumerate}[label=(\alph*)]
\item Find all values of $\theta$ in the range $0° < \theta < 360°$ satisfying
$$3\cot\theta + 4\cosec^2\theta = 5.$$ [5]

\item By writing $24\cos x - 7\sin x$ in the form $R\cos(x + \alpha)$, where $R$ and $\alpha$ are constants with $R > 0$ and $0° < \alpha < 90°$, solve the equation
$$24\cos x - 7\sin x = 16$$
for values of $x$ between $0°$ and $360°$. [6]
\end{enumerate}

\hfill \mbox{\textit{WJEC Unit 3 2024 Q2 [11]}}