3 Express the equation \(\sec \theta = 3 \cos \theta + \tan \theta\) as a quadratic equation in \(\sin \theta\). Hence solve this equation for \(- 90 ^ { \circ } < \theta < 90 ^ { \circ }\).
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Question 3:
EITHER:
Answer Marks
Guidance
Answer/Working Mark
Guidance
Correctly restate the equation in terms of \(\sin\theta\) and \(\cos\theta\) B1
Correct method to obtain a horizontal equation in \(\sin\theta\) M1
Reduce to correct quadratic, e.g. \(3\sin^2\theta - \sin\theta - 2 = 0\) A1
Solve a three-term quadratic for \(\sin\theta\) M1
Obtain final answer \(\theta = -41.8°\) only A1
Ignore answers outside the given interval
OR 1:
Answer Marks
Guidance
Answer/Working Mark
Guidance
Square both sides and use \(1 + \tan^2\theta = \sec^2\theta\) B1
Correct method to obtain a horizontal equation in \(\sin\theta\) M1
Reduce to correct quadratic, e.g. \(9\sin^2\theta - 6\sin\theta - 8 = 0\) A1
Solve a three-term quadratic for \(\sin\theta\) M1
Obtain final answer \(\theta = -41.8°\) only A1
OR 2:
Answer Marks
Guidance
Answer/Working Mark
Guidance
Multiply through by \((\sec\theta + \tan\theta)\) M1
Use \(\sec^2\theta - \tan^2\theta = 1\) B1
Obtain \(1 = 3 + 3\sin\theta\) A1
Solve for \(\sin\theta\) M1
Obtain final answer \(\theta = -41.8°\) only A1
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## Question 3:
### EITHER:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correctly restate the equation in terms of $\sin\theta$ and $\cos\theta$ | B1 | |
| Correct method to obtain a horizontal equation in $\sin\theta$ | M1 | |
| Reduce to correct quadratic, e.g. $3\sin^2\theta - \sin\theta - 2 = 0$ | A1 | |
| Solve a three-term quadratic for $\sin\theta$ | M1 | |
| Obtain final answer $\theta = -41.8°$ only | A1 | Ignore answers outside the given interval |
### OR 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Square both sides and use $1 + \tan^2\theta = \sec^2\theta$ | B1 | |
| Correct method to obtain a horizontal equation in $\sin\theta$ | M1 | |
| Reduce to correct quadratic, e.g. $9\sin^2\theta - 6\sin\theta - 8 = 0$ | A1 | |
| Solve a three-term quadratic for $\sin\theta$ | M1 | |
| Obtain final answer $\theta = -41.8°$ only | A1 | |
### OR 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Multiply through by $(\sec\theta + \tan\theta)$ | M1 | |
| Use $\sec^2\theta - \tan^2\theta = 1$ | B1 | |
| Obtain $1 = 3 + 3\sin\theta$ | A1 | |
| Solve for $\sin\theta$ | M1 | |
| Obtain final answer $\theta = -41.8°$ only | A1 | |
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3 Express the equation $\sec \theta = 3 \cos \theta + \tan \theta$ as a quadratic equation in $\sin \theta$. Hence solve this equation for $- 90 ^ { \circ } < \theta < 90 ^ { \circ }$.
\hfill \mbox{\textit{CAIE P3 2016 Q3 [5]}}