Problem Statement
Solve the equation $2\cos^2 x + 3\sin x - 3 = 0$ for $0 \le x \le 2\pi$.
**(a)** [2 marks] Show that the equation can be written as $2\sin^2 x - 3\sin x + 1 = 0$.
**(b)** [4 marks] Hence, find all solutions in the given interval in terms of $\pi$.
Verified Solution & Marking Scheme
Part (a) [M1]
Using identity $\cos^2 x = 1 - \sin^2 x$: $2(1 - \sin^2 x) + 3\sin x - 3 = 0$
Part (a) [A1]
$2 - 2\sin^2 x + 3\sin x - 3 = 0 \implies 2\sin^2 x - 3\sin x + 1 = 0$ (AG)
Part (a) [Note]
Award M1 for using the Pythagorean identity $\cos^2 x = 1 - \sin^2 x$ to form a quadratic in $\sin x$.
Part (a) [Note]
Answers must be in radians. Award A0 for angles given in degrees ($30^\circ, 90^\circ, 150^\circ$).
Part (b) [M1]
Factoring: $(2\sin x - 1)(\sin x - 1) = 0$
Part (b) [A1]
$\sin x = \frac{1}{2}$ or $\sin x = 1$
Part (b) [A1]
For $\sin x = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6}$
Part (b) [A1]
For $\sin x = 1 \implies x = \frac{\pi}{2}$