WAEC 2016 · Paper 2 · Q10

(a)
Find the range of values of $x$ for which $3x^2 + x - 2 \le 0$ .
Show the answer
$-1 \le x \le \frac23$
(b)
Solve $\cos^2\theta + \cos2\theta\sin\theta - 1 = 0$ , for $0^\circ \le \theta \le 360^\circ$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Factorise: two numbers that multiply to $3 \times (-2) = -6$ and add to 1 are 3 and $-2$ .
So $3x^2 + 3x - 2x - 2 = 3x(x + 1) - 2(x + 1)$
$= (3x - 2)(x + 1)$ , and $(3x - 2)(x + 1) \le 0$ .
The roots are $x = -1$ and $x = \frac23$ . " $\le 0$ " is between the roots, ends included: $-1 \le x \le \frac23$ .

Write everything in terms of $\sin\theta$ : $\cos^2\theta = 1 - \sin^2\theta$ and $\cos2\theta = 1 - 2\sin^2\theta$ .
Substitute: $1 - \sin^2\theta + (1 - 2\sin^2\theta)\sin\theta - 1 = 0$ .
Simplify: $-\sin^2\theta + \sin\theta - 2\sin^3\theta = 0$ .
Multiply by $-1$ : $2\sin^3\theta + \sin^2\theta - \sin\theta = 0$ .
Take out $\sin\theta$ : $\sin\theta(2\sin^2\theta + \sin\theta - 1) = 0$ .
Factorise the bracket: $\sin\theta(2\sin\theta - 1)(\sin\theta + 1) = 0$ .
$\sin\theta = 0$ gives $\theta = 0^\circ, 180^\circ, 360^\circ$ .
$\sin\theta = \frac12$ gives $\theta = 30^\circ, 150^\circ$ .
$\sin\theta = -1$ gives $\theta = 270^\circ$ .
So $\theta = 0^\circ, 30^\circ, 150^\circ, 180^\circ, 270^\circ, 360^\circ$ .