1. 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 $\tan x = \dfrac{2}{3}$ $x = -146.3^{\circ}, 33.7^{\circ}$
2. Solve, in the interval $0\leqslant\theta\leqslant360^{\circ}$: $$\tan(\theta - 45^{\circ})=-1$$
$0, 180^{\circ}, 360^{\circ}$
3. 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.
$10.2$
4. Solve, in the interval $0\leqslant\theta\leqslant180^{\circ}$: $$4\sin3\theta = 3$$ Round your answers to the nearest degree.
10. 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.
Find the exact time taken, in minutes, for the pendulum to return to its starting position.
Find all the times in the first minute that the pendulum has a displacement of $0.015$ radians.
$\dfrac{\pi}{10}$
$0.0361,0.278,0.350,0.592,0.664,0.906,0.979$
11. Solve, in the interval $0\leqslant\theta\leqslant2\pi$: $$2\sin^2\theta=3(1-\cos\theta)$$
$0, \dfrac{\pi}{3}, \dfrac{5\pi}{3}, 2\pi$
12. Solve, in the interval $0\leqslant\theta\leqslant360^{\circ}$: $$4(\sin^2\theta-\cos\theta)=3-2\cos\theta$$