Standard +0.3 This is a standard trigonometric equation requiring the identity sin²θ + cos²θ = 1 to convert to a quadratic in cosθ, then solving the resulting quadratic. It's a routine multi-step problem (substitute identity, rearrange to quadratic form, solve, find angles in range) that's slightly above average difficulty due to requiring multiple techniques, but follows a well-practiced procedure with no novel insight needed.
14 In this question you must show detailed reasoning.
Solve the equation \(5 - \cos \theta - 6 \sin ^ { 2 } \theta = 0\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).
Turn over for question 15
Using the correct identity. May be seen anywhere in working. Condone poor notation such as missing arguments.
\(6\cos^2\theta - \cos\theta - 1 = 0\)
M1
Simplifies to 3 term quadratic in cosine and sets equal to 0. Can score this mark if \(1 + \cos^2\theta\) used (gives an unsolvable quadratic).
e.g. \((3\cos\theta + 1)(2\cos\theta - 1) = 0\)
M1
Attempt to solve a quadratic in cosine. Allow sign errors only in either method. Calculator methods not acceptable here — detailed reasoning required. If using QF a correctly quoted formula followed by a slip in substitution scores M1, but if formula isn't quoted and there are errors in substitution then M0.
\(\cos\theta = \frac{1}{2}\) and \(\cos\theta = -\frac{1}{3}\) seen
A1
—
\(60°\) and \(300°\) or \(109°\) and \(251°\) or \(60°\) and \(109°\)
A1
A complete solve for either trig equation or two 'first values' obtained: allow 109.47… and 250.53…; or e.g. 109.5 and 250.5 (awrt 109 and 251)
All 4 correct with no extras
A1
SC1: If candidate uses calculator to obtain \(\cos\theta = \frac{1}{2}\) and \(\cos\theta = -\frac{1}{3}\) then proceeds to fully correct solution, award SC 4/6. SC2: If proceeds to partially correct solution with at least two correct values, award SC 3/6.
Total: [6]
## Question 14:
$5 - \cos\theta - 6(1 - \cos^2\theta) = 0$ | **B1** | Using the correct identity. May be seen anywhere in working. Condone poor notation such as missing arguments.
$6\cos^2\theta - \cos\theta - 1 = 0$ | **M1** | Simplifies to **3 term** quadratic **in cosine** and sets equal to 0. Can score this mark if $1 + \cos^2\theta$ used (gives an unsolvable quadratic).
e.g. $(3\cos\theta + 1)(2\cos\theta - 1) = 0$ | **M1** | Attempt to solve a quadratic in cosine. Allow sign errors only in either method. **Calculator methods not acceptable here — detailed reasoning required.** If using QF a correctly quoted formula followed by a slip in substitution scores M1, but if formula isn't quoted and there are errors in substitution then M0.
$\cos\theta = \frac{1}{2}$ **and** $\cos\theta = -\frac{1}{3}$ seen | **A1** | —
$60°$ **and** $300°$ or $109°$ **and** $251°$ or $60°$ **and** $109°$ | **A1** | A complete solve for either trig equation or two 'first values' obtained: allow 109.47… and 250.53…; or e.g. 109.5 and 250.5 (awrt 109 and 251)
All 4 correct with no extras | **A1** | SC1: If candidate uses calculator to obtain $\cos\theta = \frac{1}{2}$ **and** $\cos\theta = -\frac{1}{3}$ then proceeds to fully correct solution, award SC 4/6. SC2: If proceeds to partially correct solution with at least two correct values, award SC 3/6.
**Total: [6]**
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14 In this question you must show detailed reasoning.\\
Solve the equation $5 - \cos \theta - 6 \sin ^ { 2 } \theta = 0$ for $0 ^ { \circ } < \theta < 360 ^ { \circ }$.
Turn over for question 15
\hfill \mbox{\textit{OCR MEI AS Paper 2 2024 Q14 [6]}}