21M.1.SL.TZ2.3

METHOD 1

correct substitution of ${\cos }^2 x=1- {\sin }^2 x$            A1

$2\left(1-{\sin }^2 x\right)+5 \sin x=4$

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

 

METHOD 2

correct substitution using double-angle identities             A1

$\left(2 {\cos }^2 x-1\right)+5 \sin x=3$

$1-2 {\sin }^2 x-5 \sin x=3$

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

 

[1 mark]

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

 

[5 marks]

[N/A]

[N/A]