Q1
Answer
A student attempts to solve $$2\cos x = 3\sin x$$ in the interval $-180^{\circ}\leqslant x\leqslant 180^{\circ}$ as follows: $$\tan x = \dfrac{3}{2}\Rightarrow x = 56.3^{\circ}, -123.7^{\circ}$$ Identify the mistake made by the student and solve the equation.
It should be $\tan x = \dfrac{2}{3}$ <br> $x = -146.3^{\circ}, 33.7^{\circ}$
Q2
Answer
Solve, in the interval $0\leqslant\theta\leqslant360^{\circ}$: $$\tan(\theta - 45^{\circ})=-1$$
$0, 180^{\circ}, 360^{\circ}$
Q3
Answer
The height of a balloon is given by $$h = \frac{1}{2}\sin(1200t)^{\circ}$$ where $h$ is the height in metres relative to its original position, and $t$ is the time in minutes. Find the time, in seconds, at which the balloon has first fallen $0.2$ metres.
$10.2$
Q4
Answer
Solve, in the interval $0\leqslant\theta\leqslant180^{\circ}$: $$4\sin3\theta = 3$$ Round your answers to the nearest degree.
$16^{\circ}, 44^{\circ}, 136^{\circ}, 164^{\circ}$
Q5
Answer
Solve, in the interval $0\leqslant\theta\leqslant360^{\circ}$: $$2\tan\frac{1}{3}\theta = 5$$
$205^{\circ}$
Q6
Answer
Solve, in the interval $-180^{\circ}\leqslant x\leqslant180^{\circ}$: $$3\sin3x=2\cos3x$$
$-169^{\circ}, -109^{\circ}, -49^{\circ}, 11^{\circ}, 71^{\circ}, 131^{\circ}$
Q7
Answer
Solve, in the interval $-360^{\circ}\leqslant x\leqslant360^{\circ}$: $$\sqrt{3}\sin(x-60^{\circ})+\cos(x-60^{\circ})=0$$
$-330^{\circ}, -150^{\circ}, 30^{\circ}, 210^{\circ}$
Q8
Answer
Solve,in the interval $0\leqslant\theta\leqslant2\pi$: $$\tan^2\left(\theta-\frac{\pi}{4}\right)=1$$
$0, \dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}, 2\pi$
Q9
Answer
Solve, in the interval $0\leqslant\theta\leqslant360^{\circ}$: $$3\sin^2\theta+\sin\theta = 0$$
$0^{\circ}, 180^{\circ}, 199^{\circ}, 341^{\circ}, 360^{\circ}$
Q10
Answer
A pendulum is modelled by $$\theta = 0.02\cos(20t)$$ where $\theta$ is its angle of displacement relative to the vertical and $t$ is the time in minutes. All angles are measured in radians.
  1. Find the exact time taken, in minutes, for the pendulum to return to its starting position.
  2. Find all the times in the first minute that the pendulum has a displacement of $0.015$ radians.
  1. $\dfrac{\pi}{10}$
  2. $0.0361,0.278,0.350,0.592,0.664,0.906,0.979$
Q11
Answer
Solve, in the interval $0\leqslant\theta\leqslant2\pi$: $$2\sin^2\theta=3(1-\cos\theta)$$
$0, \dfrac{\pi}{3}, \dfrac{5\pi}{3}, 2\pi$
Q12
Answer
Solve, in the interval $0\leqslant\theta\leqslant360^{\circ}$: $$4(\sin^2\theta-\cos\theta)=3-2\cos\theta$$
$72^{\circ}, 144^{\circ}, 216^{\circ}, 288^{\circ}$
Q13
Answer
Solve, in the interval $-\pi \leqslant x \leqslant \pi$: $$\cos\left(2x-\dfrac{2\pi}{3}\right) = 0$$
$-\dfrac{11\pi}{12}, -\dfrac{5\pi}{12}, \dfrac{\pi}{12}, \dfrac{7\pi}{12}$
Q14
Answer
Solve, in the interval $0 \leqslant x \leqslant 2\pi$: $$\sin\left(2x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}$$
$\dfrac{\pi}{12}, \dfrac{\pi}{4}, \dfrac{13\pi}{12}, \dfrac{5\pi}{4}$
Q15
Answer
Solve, in the interval $-180^{\circ} \leqslant x \leqslant 180^{\circ}$: $$\sin(3x+45^{\circ}) = 0.5$$ Round your answers to the nearest degree.
$-125^{\circ}, -85^{\circ}, -5^{\circ}, 35^{\circ}, 115^{\circ}, 155^{\circ}$
Q16
Answer
Solve, in the interval $-\pi \leqslant x \leqslant \pi$: $$3\tan\left(\frac{\pi}{2} - 2x\right) = \sqrt{3}$$
$-\dfrac{5\pi}{6},-\dfrac{\pi}{3},\dfrac{\pi}{6},\dfrac{2\pi}{3}$
Q17
Answer
Solve, in the interval $0\leqslant\theta\leqslant360^{\circ}$: $$2\cos^22\theta-5\cos2\theta+2=0$$
$30^{\circ}, 150^{\circ}, 210^{\circ}, 330^{\circ}$
Q18
Answer
Solve, in the interval $0\leqslant\theta\leqslant180^{\circ}$: $$4\sin3\theta=\tan3\theta$$ Round your answers to the nearest degree.
$0^{\circ}, 25^{\circ}, 60^{\circ}, 94^{\circ}, 120^{\circ}, 145^{\circ}, 180^{\circ}$