18M.1.AHL.TZ1.H_8

pestleMathematicsAAHLPaper 118M· sl-3-8-solving-trig-equationssource ↗

Let $a = {\text{sin}}\,b,\,\,0 < b < \frac{\pi }{2}$ .

Find, in terms of b, the solutions of ${\text{sin}}\,2x = - a,\,\,0 \leqslant x \leqslant \pi$ .

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\text{sin}}\,2x = - {\text{sin}}\,b$

EITHER

${\text{sin}}\,2x = {\text{sin}}\left( { - b} \right)$ or ${\text{sin}}\,2x = {\text{sin}}\left( {\pi + b} \right)$ or ${\text{sin}}\,2x = {\text{sin}}\left( {2\pi - b} \right)$ … (M1)(A1)

Note: Award M1 for any one of the above, A1 for having final two.

(M1)(A1)

Note: Award M1 for one of the angles shown with b clearly labelled, A1 for both angles shown. Do not award A1 if an angle is shown in the second quadrant and subsequent A1 marks not awarded.

THEN

$2x = \pi + b$ or $2x = 2\pi - b$ (A1)(A1)

$x = \frac{\pi }{2} + \frac{b}{2},\,\,x = \pi - \frac{b}{2}$ A1

[5 marks]

Examiners’ report

[N/A]