| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: sin²/cos² substitution |
| Difficulty | Moderate -0.3 This is a standard C2 trigonometric equation requiring the identity cos²x = 1 - sin²x to convert to quadratic form (already shown in part a), then factorizing or using the quadratic formula, and finding angles in the given range. It's slightly easier than average because part (a) guides the algebraic manipulation and the quadratic factorizes neatly as (4sin x + 3)(sin x + 1) = 0, making it a routine textbook exercise with no novel problem-solving required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
\begin{enumerate}
\item (a) Show that the equation
\end{enumerate}
$$3 \sin ^ { 2 } x + 7 \sin x = \cos ^ { 2 } x - 4$$
can be written in the form
$$4 \sin ^ { 2 } x + 7 \sin x + 3 = 0$$
(b) Hence solve, for $0 \leqslant x < 360 ^ { \circ }$,
$$3 \sin ^ { 2 } x + 7 \sin x = \cos ^ { 2 } x - 4$$
giving your answers to 1 decimal place where appropriate.\\
\hfill \mbox{\textit{Edexcel C2 2011 Q7 [7]}}