| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Quadratic trigonometric equations |
| Type | Show then solve: algebraic fraction manipulation |
| Difficulty | Standard +0.3 This is a standard two-part trigonometric equation requiring routine algebraic manipulation (converting tan to sin/cos, clearing fractions, using sin²+cos²=1) followed by solving a quadratic in sin x. The steps are methodical and commonly practiced in P1, making it slightly easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tan x + \cos x = k(\tan x - \cos x)\) leading to \(\sin x + \cos^2 x = k(\sin x - \cos^2 x)\) | M1 | Use \(\tan x = \frac{\sin x}{\cos x}\) and clear fraction. |
| \(\sin x + 1 - \sin^2 x = k\sin x - k + k\sin^2 x\) | \*M1 | Use \(\cos^2 x = 1 - \sin^2 x\) twice to obtain an equation in sine. |
| \(k\sin^2 x + \sin^2 x + k\sin x - \sin x - k - 1 = 0\) | DM1 | Gather like terms on one side of the equation. |
| \((k+1)\sin^2 x + (k-1)\sin x - (k+1) = 0\) | A1 | AG. Factorise to obtain answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(5\sin^2 x + 3\sin x - 5 = 0\) | B1 | |
| \(\sin x = \dfrac{-3 \pm \sqrt{9 + 100}}{10}\) | M1 | Use formula or complete the square. |
| \(x = 48.1°, \ 131.9°\) | A1, A1 FT | AWRT. Maximum A1 if extra solutions in range. FT for \(180 - \text{their}\) answer or \(540 - \text{their}\) answer if \(\sin x\) is negative. If M0 given and correct answers only SCB1B1 available. If answers in radians; \(0.839, 2.30\) can score SCB1 for both. |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan x + \cos x = k(\tan x - \cos x)$ leading to $\sin x + \cos^2 x = k(\sin x - \cos^2 x)$ | M1 | Use $\tan x = \frac{\sin x}{\cos x}$ and clear fraction. |
| $\sin x + 1 - \sin^2 x = k\sin x - k + k\sin^2 x$ | \*M1 | Use $\cos^2 x = 1 - \sin^2 x$ twice to obtain an equation in sine. |
| $k\sin^2 x + \sin^2 x + k\sin x - \sin x - k - 1 = 0$ | DM1 | Gather like terms on one side of the equation. |
| $(k+1)\sin^2 x + (k-1)\sin x - (k+1) = 0$ | A1 | AG. Factorise to obtain answer. |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5\sin^2 x + 3\sin x - 5 = 0$ | B1 | |
| $\sin x = \dfrac{-3 \pm \sqrt{9 + 100}}{10}$ | M1 | Use formula or complete the square. |
| $x = 48.1°, \ 131.9°$ | A1, A1 FT | AWRT. Maximum A1 if extra solutions in range. FT for $180 - \text{their}$ answer or $540 - \text{their}$ answer if $\sin x$ is negative. If M0 given and correct answers only **SCB1B1** available. If answers in radians; $0.839, 2.30$ can score **SCB1** for both. |7
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\frac { \tan x + \cos x } { \tan x - \cos x } = k$, where $k$ is a constant, can be expressed as
$$( k + 1 ) \sin ^ { 2 } x + ( k - 1 ) \sin x - ( k + 1 ) = 0$$
\item Hence solve the equation $\frac { \tan x + \cos x } { \tan x - \cos x } = 4$ for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q7 [8]}}