(i) Show that \(\sin^4 \theta - \cos^4 \theta \equiv 2 \sin^2 \theta - 1\).
(ii) Hence solve the equation \(\sin^4 \theta - \cos^4 \theta = \frac{1}{2}\) for \(0^\circ \leq \theta \leq 360^\circ\).
Solve the equation \(\frac{13 \sin^2 \theta}{2 + \cos \theta} + \cos \theta = 2\) for \(0^\circ \leq \theta \leq 180^\circ\).
(i) Prove the identity \(\frac{\tan x + 1}{\sin x \tan x + \cos x} \equiv \sin x + \cos x\).
(ii) Hence solve the equation \(\frac{\tan x + 1}{\sin x \tan x + \cos x} = 3 \sin x - 2 \cos x\) for \(0 \leq x \leq 2\pi\).
(i) Prove the identity \(\frac{1}{\cos \theta} - \frac{\cos \theta}{1 + \sin \theta} \equiv \tan \theta\).
(ii) Solve the equation \(\frac{1}{\cos \theta} - \frac{\cos \theta}{1 + \sin \theta} + 2 = 0\) for \(0^\circ \leq \theta \leq 360^\circ\).
(i) Prove the identity \(\frac{\sin \theta}{1 - \cos \theta} - \frac{1}{\sin \theta} \equiv \frac{1}{\tan \theta}\).
(ii) Hence solve the equation \(\frac{\sin \theta}{1 - \cos \theta} - \frac{1}{\sin \theta} = 4 \tan \theta\) for \(0^\circ < \theta < 180^\circ\).
(i) Show that \(\frac{\sin \theta}{\sin \theta + \cos \theta} + \frac{\cos \theta}{\sin \theta - \cos \theta} \equiv \frac{1}{\sin^2 \theta - \cos^2 \theta}\).
(ii) Hence solve the equation \(\frac{\sin \theta}{\sin \theta + \cos \theta} + \frac{\cos \theta}{\sin \theta - \cos \theta} = 3\), for \(0^\circ \leq \theta \leq 360^\circ\).
Solve, by factorising, the equation
\(6 \cos \theta \tan \theta - 3 \cos \theta + 4 \tan \theta - 2 = 0,\)
for \(0^\circ \leq \theta \leq 180^\circ\).
(i) Solve the equation \(\sin 2x + 3 \cos 2x = 0\) for \(0^\circ \leq x \leq 360^\circ\).
(ii) How many solutions has the equation \(\sin 2x + 3 \cos 2x = 0\) for \(0^\circ \leq x \leq 1080^\circ\)?
Solve the equation \(\sin 2x = 2 \cos 2x\), for \(0^\circ \leq x \leq 180^\circ\).
(i) Prove the identity \(\left( \frac{1}{\sin \theta} - \frac{1}{\tan \theta} \right)^2 \equiv \frac{1 - \cos \theta}{1 + \cos \theta}\).
(ii) Hence solve the equation \(\left( \frac{1}{\sin \theta} - \frac{1}{\tan \theta} \right)^2 = \frac{2}{5}\), for \(0^\circ \leq \theta \leq 360^\circ\).
(i) Prove the identity \(\frac{\cos \theta}{\tan \theta (1 - \sin \theta)} \equiv 1 + \frac{1}{\sin \theta}\).
(ii) Hence solve the equation \(\frac{\cos \theta}{\tan \theta (1 - \sin \theta)} = 4\), for \(0^\circ \leq \theta \leq 360^\circ\).
(i) Prove the identity \(\frac{\sin x \tan x}{1 - \cos x} \equiv 1 + \frac{1}{\cos x}\).
(ii) Hence solve the equation \(\frac{\sin x \tan x}{1 - \cos x} + 2 = 0\), for \(0^\circ \leq x \leq 360^\circ\).
(i) Show that the equation
\(3(2 \sin x - \cos x) = 2(\sin x - 3 \cos x)\)
can be written in the form \(\tan x = -\frac{3}{4}\).
(ii) Solve the equation \(3(2 \sin x - \cos x) = 2(\sin x - 3 \cos x)\), for \(0^\circ \leq x \leq 360^\circ\).
(i) Prove the identity \((\sin x + \cos x)(1 - \sin x \cos x) \equiv \sin^3 x + \cos^3 x\).
(ii) Solve the equation \((\sin x + \cos x)(1 - \sin x \cos x) = 9 \sin^3 x\) for \(0^\circ \leq x < 360^\circ\).
Solve the equation \(3 \tan(2x + 15^\circ) = 4\) for \(0^\circ \leq x \leq 180^\circ\).
Solve the equation \(\sin 2x + 3 \cos 2x = 0\), for \(0^\circ \leq x < 180^\circ\).
(i) Show that the equation \(\sin \theta + \cos \theta = 2(\sin \theta - \cos \theta)\) can be expressed as \(\tan \theta = 3\).
(ii) Hence solve the equation \(\sin \theta + \cos \theta = 2(\sin \theta - \cos \theta)\), for \(0^\circ \leq \theta \leq 360^\circ\).
(a) Show that the equation \(\frac{\tan x + \sin x}{\tan x - \sin x} = k\), where \(k\) is a constant, may be expressed as \(\frac{1 + \cos x}{1 - \cos x} = k\).
(b) Hence express \(\cos x\) in terms of \(k\).
(c) Hence solve the equation \(\frac{\tan x + \sin x}{\tan x - \sin x} = 4\) for \(-\pi < x < \pi\).
Find all the values of \(x\) in the interval \(0^\circ \leq x \leq 180^\circ\) which satisfy the equation \(\sin 3x + 2 \cos 3x = 0\).
(a) Prove the identity \(\frac{1 + \sin x}{1 - \sin x} - \frac{1 - \sin x}{1 + \sin x} \equiv \frac{4 \tan x}{\cos x}\).
(b) Hence solve the equation \(\frac{1 + \sin x}{1 - \sin x} - \frac{1 - \sin x}{1 + \sin x} = 8 \tan x\) for \(0 \leq x \leq \frac{1}{2} \pi\).