Edexcel C2 — Question 4 7 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
Marks7
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TopicQuadratic trigonometric equations
TypeDirect solve: sin²/cos² substitution
DifficultyModerate -0.3 This is a standard C2 trigonometric equation requiring the identity sin²θ = 1 - cos²θ to convert to a quadratic in cos θ, then solving the resulting quadratic. It's slightly easier than average because it's a routine textbook exercise with a clear method and no conceptual surprises, though it does require multiple steps (substitution, quadratic formula, inverse trig).
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
4. Solve the equation $$\sin ^ { 2 } \theta = 4 \cos \theta ,$$ for values of \(\theta\) in the interval \(0 \leq \theta \leq 360 ^ { \circ }\).
AnswerMarks
\(1 - \cos^2 \theta = 4 \cos \theta\)M1
\(\cos^2 \theta + 4 \cos \theta - 1 = 0\)A1
\(\cos \theta = \frac{-4 \pm \sqrt{16+4}}{2} = -2 - \sqrt{5}\) (no solutions) or \(-2 + \sqrt{5}\)M1 A1
\(\theta = 76.3, 360 - 76.3\)B1 M1
\(\theta = 76.3°, 283.7°\) (1dp)A1 
$1 - \cos^2 \theta = 4 \cos \theta$ | M1 |
$\cos^2 \theta + 4 \cos \theta - 1 = 0$ | A1 |
$\cos \theta = \frac{-4 \pm \sqrt{16+4}}{2} = -2 - \sqrt{5}$ (no solutions) or $-2 + \sqrt{5}$ | M1 A1 |
$\theta = 76.3, 360 - 76.3$ | B1 M1 |
$\theta = 76.3°, 283.7°$ (1dp) | A1 |

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4. Solve the equation

$$\sin ^ { 2 } \theta = 4 \cos \theta ,$$

for values of $\theta$ in the interval $0 \leq \theta \leq 360 ^ { \circ }$.\\

\hfill \mbox{\textit{Edexcel C2  Q4 [7]}}
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