OCR C4 — Question 6 7 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeDouble angle equations requiring identity expansion and factorisation
DifficultyStandard +0.3 This question requires applying the double angle formula cos 2θ = 1 - 2sin²θ, then solving a quadratic equation in sin θ, followed by finding angles in the given range. It's a standard C4 trigonometric identity and equation question with straightforward steps, making it slightly easier than average.
Spec1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals
Express \(6\cos 2\theta + \sin\theta\) in terms of \(\sin\theta\). Hence solve the equation \(6\cos 2\theta + \sin\theta = 0\), for \(0° \leqslant \theta \leqslant 360°\). [7]
Question 6:
AnswerMarks
6cos212sin2
(6cos2sin) 612sin2sin
6cos2+ sin= 0
 12sin2 − sin − 6 = 0
 (4sin − 3)(3sin + 2) = 0
 sin = 3/4 or −2/3
 sin = 3/4,  = 48.6°, 131.4°
AnswerMarks
sin = −2/3,  = 221.8°, 318.2°M1*
A1
M1dep*
A1
B1
B1
B1
AnswerMarks
[7]cos212sin2(maybe implied in substitution)
Use of correct quadratic equation formula or factorising or comp. the
square on their three term quadratic in sin (see guidance in question
1 for awarding this method mark) providedb24ac0
www
First correct solution to 1 dp or better (eg 48.59° etc)
Three correct solutions
All four correct solutions and no others in the range
Ignore solutions outside the range
SC Award max B1B1B0 for answers in radians (0.85, 2.29, 3.87, 5.55
or better – so one correct B1, three correct B1). Award max B1 if
there are extra solutions in the range with radians
SC If M1M1 awarded and both values of sin1but B0B0B0 then
award B1 only for evidence of usingsinsin  180
Question 6:
6 | cos212sin2
(6cos2sin) 612sin2sin
6cos2+ sin= 0
 12sin2 − sin − 6 = 0
 (4sin − 3)(3sin + 2) = 0
 sin = 3/4 or −2/3
 sin = 3/4,  = 48.6°, 131.4°
sin = −2/3,  = 221.8°, 318.2° | M1*
A1
M1dep*
A1
B1
B1
B1
[7] | cos212sin2(maybe implied in substitution)
Use of correct quadratic equation formula or factorising or comp. the
square on their three term quadratic in sin (see guidance in question
1 for awarding this method mark) providedb24ac0
www
First correct solution to 1 dp or better (eg 48.59° etc)
Three correct solutions
All four correct solutions and no others in the range
Ignore solutions outside the range
SC Award max B1B1B0 for answers in radians (0.85, 2.29, 3.87, 5.55
or better – so one correct B1, three correct B1). Award max B1 if
there are extra solutions in the range with radians
SC If M1M1 awarded and both values of sin1but B0B0B0 then
award B1 only for evidence of usingsinsin  180
Express $6\cos 2\theta + \sin\theta$ in terms of $\sin\theta$.

Hence solve the equation $6\cos 2\theta + \sin\theta = 0$, for $0° \leqslant \theta \leqslant 360°$. [7]

\hfill \mbox{\textit{OCR C4  Q6 [7]}}
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