Clavis TMUA Set A Paper 2 · Question 13

Consider the following attempt to solve the equation $2\sin x\cos x=\cos x$ for $0\leq x\leq2\pi$ .

(I) Dividing both sides by $\cos x$ gives $2\sin x=1$ .
(II) So $\sin x=\frac12$ .
(III) The solutions of $\sin x=\frac12$ with $0\leq x\leq2\pi$ are $x=\frac\pi6$ and $x=\frac{5\pi}6$ .
(IV) Therefore the solutions of the original equation are $x=\frac\pi6$ and $x=\frac{5\pi}6$ .

Which one of the following is true?

AThe working is correct.

BThe two values found are not solutions of the original equation, and the error arises as a result of step (I).

CThe two values found are solutions, but there is exactly one further solution, missed as a result of step (I).

DThe two values found are solutions, but there are exactly two further solutions, missed as a result of step (I).

EThe two values found are solutions, but there are exactly two further solutions, missed as a result of step (III).

FThe two values found are solutions, but there are exactly four further solutions, missed as a result of step (I).

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