21M.1.AHL.TZ2.2

attempt to use ${\cos }^2 x=1-{\sin }^2 x$              M1

$2 {\sin }^2 x-5 \sin x+2=0$                   A1

 

EITHER

attempting to factorise              M1

$(2 \sin x-1)(\sin x-2)$                   A1

 

OR

attempting to use the quadratic formula            M1

$\sin x=\frac{5\pm \sqrt{5^2-4\times 2\times 2}}{4}\left(=\frac{5\pm 3}{4}\right)$         A1

 

THEN

$\sin x=\frac{1}{2}$           (A1)

$x=\frac{\mathrm{\pi }}{6}, \frac{5\mathrm{\pi }}{6}$                  A1A1

 

[7 marks]

[N/A]