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 cos²x = 1 - sin²x to convert to a quadratic in sin x, then solving the quadratic and finding angles in the given range. It's slightly easier than average as it follows a well-practiced procedure with no conceptual surprises, though it does require multiple steps including quadratic formula and considering all solutions in the domain.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
4. Solve, for \(0 \leq x < 360\), the equation $$3 \cos ^ { 2 } x ^ { \circ } + \sin ^ { 2 } x ^ { \circ } + 5 \sin x ^ { \circ } = 0$$
AnswerMarks Guidance
\(3(1 - \sin^2 x) + \sin^2 x + 5\sin x = 0\)M1
\(2\sin^2 x - 5\sin x - 3 = 0\)A1
\((2\sin x + 1)(\sin x - 3) = 0\)M1
\(\sin x = 3\) (no solutions) or \(\sin x = -\frac{1}{2}\)A1
\(x = 180 + 30, 360 - 30\)B1 M1
\(x = 210, 330\)A1 (7 marks) 
$3(1 - \sin^2 x) + \sin^2 x + 5\sin x = 0$ | M1 |
$2\sin^2 x - 5\sin x - 3 = 0$ | A1 |
$(2\sin x + 1)(\sin x - 3) = 0$ | M1 |
$\sin x = 3$ (no solutions) or $\sin x = -\frac{1}{2}$ | A1 |
$x = 180 + 30, 360 - 30$ | B1 M1 |
$x = 210, 330$ | A1 | **(7 marks)**
4. Solve, for $0 \leq x < 360$, the equation

$$3 \cos ^ { 2 } x ^ { \circ } + \sin ^ { 2 } x ^ { \circ } + 5 \sin x ^ { \circ } = 0$$

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