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

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

\({\text{sin}}\,2x =  - {\text{sin}}\,b\)

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.

OR

     (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]