Solve the equation \(4 \sin \theta + \tan \theta = 0\) for \(0^\circ < \theta < 180^\circ\).

(a) Prove the identity \(\frac{1 + \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 + \sin \theta} \equiv \frac{2}{\cos \theta}\).

(b) Hence solve the equation \(\frac{1 + \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 + \sin \theta} = \frac{3}{\sin \theta}\), for \(0 \leq \theta \leq 2\pi\).

Solve the equation

\(\frac{\tan \theta + 3 \sin \theta + 2}{\tan \theta - 3 \sin \theta + 1} = 2\)

for \(0^\circ \leq \theta \leq 90^\circ\).

Given that \(x > 0\), find the two smallest values of \(x\), in radians, for which \(3 \tan(2x + 1) = 1\). Show all necessary working.

(i) Prove the identity \(\left( \frac{1}{\cos x} - \tan x \right)^2 \equiv \frac{1 - \sin x}{1 + \sin x}\).

(ii) Hence solve the equation \(\left( \frac{1}{\cos 2x} - \tan 2x \right)^2 = \frac{1}{3}\) for \(0 \leq x \leq \pi\).

(i) Express \(\frac{\tan^2 \theta - 1}{\tan^2 \theta + 1}\) in the form \(a \sin^2 \theta + b\), where \(a\) and \(b\) are constants to be found.

(ii) Hence, or otherwise, and showing all necessary working, solve the equation \(\frac{\tan^2 \theta - 1}{\tan^2 \theta + 1} = \frac{1}{4}\) for \(-90^\circ \leq \theta \leq 0^\circ\).

(i) Prove the identity \((\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) \equiv \sin^3 \theta + \cos^3 \theta\).

(ii) Hence solve the equation \((\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) = 3 \cos^3 \theta\) for \(0^\circ \leq \theta \leq 360^\circ\).

(i) Show that the equation \(\frac{2 \sin \theta + \cos \theta}{\sin \theta + \cos \theta} = 2 \tan \theta\) may be expressed as \(\cos^2 \theta = 2 \sin^2 \theta\).

(ii) Hence solve the equation \(\frac{2 \sin \theta + \cos \theta}{\sin \theta + \cos \theta} = 2 \tan \theta\) for \(0^\circ < \theta < 180^\circ\).

(i) Prove the identity \(\left( \frac{1}{\cos \theta} - \tan \theta \right)^2 \equiv \frac{1 - \sin \theta}{1 + \sin \theta}\).

(ii) Hence solve the equation \(\left( \frac{1}{\cos \theta} - \tan \theta \right)^2 = \frac{1}{2}\), for \(0^\circ \leq \theta \leq 360^\circ\).

(i) Prove the identity \(\frac{1 + \cos \theta}{\sin \theta} + \frac{\sin \theta}{1 + \cos \theta} \equiv \frac{2}{\sin \theta}\).

(ii) Hence solve the equation \(\frac{1 + \cos \theta}{\sin \theta} + \frac{\sin \theta}{1 + \cos \theta} = \frac{3}{\cos \theta}\) for \(0^\circ \leq \theta \leq 360^\circ\).

Showing all necessary working, solve the equation \(6 \sin^2 x - 5 \cos^2 x = 2 \sin^2 x + \cos^2 x\) for \(0^\circ \leq x \leq 360^\circ\).

Find the exact solutions of the equation \(4 \sin\left(\frac{1}{2}x - 30^\circ\right) = 2\sqrt{2}\) for \(0^\circ \leq x \leq 360^\circ\).

(i) Express the equation \(\sin 2x + 3 \cos 2x = 3(\sin 2x - \cos 2x)\) in the form \(\tan 2x = k\), where \(k\) is a constant.

(ii) Hence solve the equation for \(-90^\circ \leq x \leq 90^\circ\).

(i) Prove the identity \(\frac{1 + \cos \theta}{1 - \cos \theta} - \frac{1 - \cos \theta}{1 + \cos \theta} \equiv \frac{4}{\sin \theta \tan \theta}\).

(ii) Hence solve, for \(0^\circ < \theta < 360^\circ\), the equation \(\sin \theta \left( \frac{1 + \cos \theta}{1 - \cos \theta} - \frac{1 - \cos \theta}{1 + \cos \theta} \right) = 3\).

(i) Prove the identity \(\left( \frac{1}{\sin x} - \frac{1}{\tan x} \right)^2 \equiv \frac{1 - \cos x}{1 + \cos x}\).

(ii) Hence solve the equation \(\left( \frac{1}{\sin x} - \frac{1}{\tan x} \right)^2 = \frac{2}{5}\) for \(0 \leq x \leq 2\pi\).

(i) Express the equation \(3 \sin \theta = \cos \theta\) in the form \(\tan \theta = k\) and solve the equation for \(0^\circ < \theta < 180^\circ\).

(ii) Solve the equation \(3 \sin^2 2x = \cos^2 2x\) for \(0^\circ < x < 180^\circ\).

(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 \(0 \leq x \leq \frac{1}{2} \pi\).

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 \(0^\circ < \theta < 180^\circ\).