Solve the equation \(\frac{\tan \theta + 2 \sin \theta}{\tan \theta - 2 \sin \theta} = 3\) for \(0^\circ < \theta < 180^\circ\).
(a) Prove the identity \(\left( \frac{1}{\cos x} - \tan x \right) \left( \frac{1}{\sin x} + 1 \right) \equiv \frac{1}{\tan x}\).
(b) Hence solve the equation \(\left( \frac{1}{\cos x} - \tan x \right) \left( \frac{1}{\sin x} + 1 \right) = 2 \tan^2 x\) for \(0^\circ \leq x \leq 180^\circ\).
(a) Show that \(\frac{\sin \theta}{1 - \sin \theta} - \frac{\sin \theta}{1 + \sin \theta} \equiv 2 \tan^2 \theta\).
(b) Hence solve the equation \(\frac{\sin \theta}{1 - \sin \theta} - \frac{\sin \theta}{1 + \sin \theta} = 8\), for \(0^\circ < \theta < 180^\circ\).
(a) Show that \(\frac{\tan \theta}{1 + \cos \theta} + \frac{\tan \theta}{1 - \cos \theta} \equiv \frac{2}{\sin \theta \cos \theta}\).
(b) Hence solve the equation \(\frac{\tan \theta}{1 + \cos \theta} + \frac{\tan \theta}{1 - \cos \theta} = \frac{6}{\tan \theta}\) for \(0^\circ < \theta < 180^\circ\).
(a) Show that the equation
\(5 \cos \theta - \sin \theta \tan \theta + 1 = 0\)
may be expressed in the form \(a \cos^2 \theta + b \cos \theta + c = 0\), where \(a, b\) and \(c\) are constants to be found.
(b) Hence solve the equation \(5 \cos \theta - \sin \theta \tan \theta + 1 = 0\) for \(0 < \theta < 2\pi\).
(a) Show that the equation \(\frac{1}{\sin \theta + \cos \theta} + \frac{1}{\sin \theta - \cos \theta} = 1\) may be expressed in the form \(a \sin^2 \theta + b \sin \theta + c = 0\), where \(a, b\) and \(c\) are constants to be found.
(b) Hence solve the equation \(\frac{1}{\sin \theta + \cos \theta} + \frac{1}{\sin \theta - \cos \theta} = 1\) for \(0^\circ \leq \theta \leq 360^\circ\).
(a) Solve the equation \(6\sqrt{y} + \frac{2}{\sqrt{y}} - 7 = 0\).
(b) Hence solve the equation \(6\sqrt{\tan x} + \frac{2}{\sqrt{\tan x}} - 7 = 0\) for \(0^\circ \leq x \leq 360^\circ\).
The function \(f\) is given by \(f(x) = 4 \cos^4 x + \cos^2 x - k\) for \(0 \leq x \leq 2\pi\), where \(k\) is a constant.
(a) Given that \(k = 3\), find the exact solutions of the equation \(f(x) = 0\).
(b) Use the quadratic formula to show that, when \(k > 5\), the equation \(f(x) = 0\) has no solutions.
(a) Prove the identity \(\frac{\sin^3 \theta}{\sin \theta - 1} - \frac{\sin^2 \theta}{1 + \sin \theta} \equiv -\tan^2 \theta (1 + \sin^2 \theta)\).
(b) Hence solve the equation \(\frac{\sin^3 \theta}{\sin \theta - 1} - \frac{\sin^2 \theta}{1 + \sin \theta} = \tan^2 \theta (1 - \sin^2 \theta)\) for \(0 < \theta < 2\pi\).
(a) Show that \(\frac{\sin \theta + 2 \cos \theta}{\cos \theta - 2 \sin \theta} - \frac{\sin \theta - 2 \cos \theta}{\cos \theta + 2 \sin \theta} \equiv \frac{4}{5 \cos^2 \theta - 4}\).
(b) Hence solve the equation \(\frac{\sin \theta + 2 \cos \theta}{\cos \theta - 2 \sin \theta} - \frac{\sin \theta - 2 \cos \theta}{\cos \theta + 2 \sin \theta} = 5\) for \(0^\circ < \theta < 180^\circ\).
(a) Show that the equation \(\frac{\tan x + \cos x}{\tan x - \cos x} = k\), where \(k\) is a constant, can be expressed as \((k + 1) \sin^2 x + (k - 1) \sin x - (k + 1) = 0\).
(b) Hence solve the equation \(\frac{\tan x + \cos x}{\tan x - \cos x} = 4\) for \(0^\circ \leq x \leq 360^\circ\).
Solve the equation \(2 \cos \theta = 7 - \frac{3}{\cos \theta}\) for \(-90^\circ < \theta < 90^\circ\).
(a) Prove the identity \(\frac{1 - 2 \sin^2 \theta}{1 - \sin^2 \theta} \equiv 1 - \tan^2 \theta\).
(b) Hence solve the equation \(\frac{1 - 2 \sin^2 \theta}{1 - \sin^2 \theta} = 2 \tan^4 \theta\) for \(0^\circ \leq \theta \leq 180^\circ\).
Solve the equation \(3 \tan^2 \theta + 1 = \frac{2}{\tan^2 \theta}\) for \(0^\circ < \theta < 180^\circ\).
(i) Show that the equation \(3 \cos^4 \theta + 4 \sin^2 \theta - 3 = 0\) can be expressed as \(3x^2 - 4x + 1 = 0\), where \(x = \cos^2 \theta\).
(ii) Hence solve the equation \(3 \cos^4 \theta + 4 \sin^2 \theta - 3 = 0\) for \(0^\circ \leq \theta \leq 180^\circ\).
(a) Verify the identity \((2x - 1)(4x^2 + 2x - 1) \equiv 8x^3 - 4x + 1\).
(b) Prove the identity \(\frac{\tan^2 \theta + 1}{\tan^2 \theta - 1} \equiv \frac{1}{1 - 2 \cos^2 \theta}\).
(c) Using the results of (a) and (b), solve the equation \(\frac{\tan^2 \theta + 1}{\tan^2 \theta - 1} = 4 \cos \theta\), for \(0^\circ \leq \theta \leq 180^\circ\).
(i) Given that \(4 \tan x + 3 \cos x + \frac{1}{\cos x} = 0\), show, without using a calculator, that \(\sin x = -\frac{2}{3}\).
(ii) Hence, showing all necessary working, solve the equation \(4 \tan(2x - 20^\circ) + 3 \cos(2x - 20^\circ) + \frac{1}{\cos(2x - 20^\circ)} = 0\) for \(0^\circ \leq x \leq 180^\circ\).
Solve the equation \(3 \sin^2 2\theta + 8 \cos 2\theta = 0\) for \(0^\circ \leq \theta \leq 180^\circ\).
(i) Show that \(\frac{\tan \theta + 1}{1 + \cos \theta} + \frac{\tan \theta - 1}{1 - \cos \theta} \equiv \frac{2(\tan \theta - \cos \theta)}{\sin^2 \theta}\).
(ii) Hence, showing all necessary working, solve the equation \(\frac{\tan \theta + 1}{1 + \cos \theta} + \frac{\tan \theta - 1}{1 - \cos \theta} = 0\) for \(0^\circ < \theta < 90^\circ\).
(i) Show that the equation \(\frac{\cos \theta - 4}{\sin \theta} - \frac{4 \sin \theta}{5 \cos \theta - 2} = 0\) may be expressed as \(9 \cos^2 \theta - 22 \cos \theta + 4 = 0\).
(ii) Hence solve the equation \(\frac{\cos \theta - 4}{\sin \theta} - \frac{4 \sin \theta}{5 \cos \theta - 2} = 0\) for \(0^\circ \leq \theta \leq 360^\circ\).