Question 1Easy3 marks
A student is asked to solve the equation \(2\sin x\cos x = \sin x\) for \(0 \le x \lt 360^\circ\). The student writes:
Dividing both sides by \(\sin x\): \(2\cos x = 1\), so \(\cos x = \frac{1}{2}\) and \(x = 60^\circ\) or \(x = 300^\circ\).
Dividing both sides by \(\sin x\): \(2\cos x = 1\), so \(\cos x = \frac{1}{2}\) and \(x = 60^\circ\) or \(x = 300^\circ\).
(a) Explain the error in the student's method.[1]
(b) Find all the solutions of the equation for \(0 \le x \lt 360^\circ\).[2]
Show the answer and mark scheme
(a) Answer: Dividing by \(\sin x\) loses the solutions where \(\sin x = 0\).
- B1 for dividing by \(\sin x\) loses the solutions with \(\sin x = 0\) (you cannot divide by zero)
Worked solution: The student divided by \(\sin x\), which is not allowed when \(\sin x = 0\). The solutions with \(\sin x = 0\) have been lost.
(b) Answer: \(x = 0^\circ, 60^\circ, 180^\circ, 300^\circ\)
- M1 for factorising: \(\sin x(2\cos x - 1) = 0\)
- A1 for \(0^\circ, 60^\circ, 180^\circ, 300^\circ\) and no others
Worked solution: \(2\sin x\cos x - \sin x = 0 \Rightarrow \sin x(2\cos x - 1) = 0\).
\(\sin x = 0\): \(x = 0^\circ, 180^\circ\). \(\cos x = \frac{1}{2}\): \(x = 60^\circ, 300^\circ\).