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]