AQA C4 2012 June — Question 2 7 marks

Exam BoardAQA
ModuleC4 (Core Mathematics 4)
Year2012
SessionJune
Marks7
PaperDownload PDF ↗
TopicHarmonic Form
TypeExpress and solve equation
DifficultyStandard +0.3 This is a standard two-part harmonic form question requiring routine application of R-formula (finding R and α using Pythagorean identity and tan), then solving a simple equation. It's slightly above average difficulty due to the multi-step nature and need for careful angle work, but follows a well-practiced textbook procedure with no novel insight required.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals

2
  1. Express \(\sin x - 3 \cos x\) in the form \(R \sin ( x - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving your value of \(\alpha\) to the nearest \(0.1 ^ { \circ }\).
  2. Hence find the values of \(x\) in the interval \(0 ^ { \circ } < x < 360 ^ { \circ }\) for which $$\sin x - 3 \cos x + 2 = 0$$ giving your values of \(x\) to the nearest degree.

2
\begin{enumerate}[label=(\alph*)]
\item Express $\sin x - 3 \cos x$ in the form $R \sin ( x - \alpha )$, where $R > 0$ and $0 ^ { \circ } < \alpha < 90 ^ { \circ }$, giving your value of $\alpha$ to the nearest $0.1 ^ { \circ }$.
\item Hence find the values of $x$ in the interval $0 ^ { \circ } < x < 360 ^ { \circ }$ for which

$$\sin x - 3 \cos x + 2 = 0$$

giving your values of $x$ to the nearest degree.
\end{enumerate}

\hfill \mbox{\textit{AQA C4 2012 Q2 [7]}}