Standard +0.3 This is a standard C2 trigonometric equation requiring the identity cos²x = 1 - sin²x to convert to a quadratic in sin x, then solving the quadratic and finding angles. It's slightly above average difficulty due to the multi-step process and need to find all solutions in the given interval, but follows a well-practiced routine with no novel insight required.
6. Find all the solutions, in the interval \(0 \leqslant x < 2 \pi\), of the equation
$$2 \cos ^ { 2 } x + 1 = 5 \sin x$$
giving each solution in terms of \(\pi\).
Use of \(\cos^2 x = 1 - \sin^2 x\). Condone invisible brackets if \(2 - 2\sin^2 x\) is present or implied
\(2\sin^2 x + 5\sin x - 3 = 0\)
Must be using \(\cos^2 x = 1 - \sin^2 x\); using \(\cos^2 x = 1 + \sin^2 x\) is M0
\((2\sin x - 1)(\sin x + 3) = 0\)
\(\sin x = \frac{1}{2}\)
M1, A1
Attempt to solve quadratic in \(\sin x\). Correct factorising and \(\sin x = \frac{1}{2}\). Note \((\sin x + 3)\) as factor \(\rightarrow \sin x = -3\) can be ignored
\(x = \frac{\pi}{6}, \frac{5\pi}{6}\)
M1, M1, A1cso (6)
First M1: method for finding any angle in any range consistent with trig equation. Second M1: method for finding second angle in radians, range \(0 \leq x < 2\pi\), must involve using \(\pi\). A1: \(\frac{\pi}{6}, \frac{5\pi}{6}\) c.s.o. Recurring decimals okay instead of \(\frac{1}{6}\) and \(\frac{5}{6}\)
- Answer only \(\frac{\pi}{6}\): M0, M0, A0, M1, M0, A0
- Finding answers by trying different values (e.g. multiples of \(\pi\)): as for answer only
- Ignore extra solutions outside range; deduct final A mark for extra solutions in range
## Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2(1 - \sin^2 x) + 1 = 5\sin x$ | M1 | Use of $\cos^2 x = 1 - \sin^2 x$. Condone invisible brackets if $2 - 2\sin^2 x$ is present or implied |
| $2\sin^2 x + 5\sin x - 3 = 0$ | | Must be using $\cos^2 x = 1 - \sin^2 x$; using $\cos^2 x = 1 + \sin^2 x$ is M0 |
| $(2\sin x - 1)(\sin x + 3) = 0$ | | |
| $\sin x = \frac{1}{2}$ | M1, A1 | Attempt to solve quadratic in $\sin x$. Correct factorising and $\sin x = \frac{1}{2}$. Note $(\sin x + 3)$ as factor $\rightarrow \sin x = -3$ can be ignored |
| $x = \frac{\pi}{6}, \frac{5\pi}{6}$ | M1, M1, A1cso **(6)** | First M1: method for finding any angle in any range consistent with trig equation. Second M1: method for finding second angle in radians, range $0 \leq x < 2\pi$, must involve using $\pi$. A1: $\frac{\pi}{6}, \frac{5\pi}{6}$ c.s.o. Recurring decimals okay instead of $\frac{1}{6}$ and $\frac{5}{6}$ |
**Special cases:**
- Answer only $\frac{\pi}{6}, \frac{5\pi}{6}$: M0, M0, A0, M1, M1, A1
- Answer only $\frac{\pi}{6}$: M0, M0, A0, M1, M0, A0
- Finding answers by trying different values (e.g. multiples of $\pi$): as for answer only
- Ignore extra solutions outside range; deduct final A mark for extra solutions in range
---
6. Find all the solutions, in the interval $0 \leqslant x < 2 \pi$, of the equation
$$2 \cos ^ { 2 } x + 1 = 5 \sin x$$
giving each solution in terms of $\pi$.\\
\hfill \mbox{\textit{Edexcel C2 2007 Q6 [6]}}