Start with the equation \(3 \tan(2x + 15^\circ) = 4\).

Divide both sides by 3 to isolate the tangent function:

\(\tan(2x + 15^\circ) = \frac{4}{3}\).

Find the angle whose tangent is \(\frac{4}{3}\):

\(2x + 15^\circ = \tan^{-1}\left(\frac{4}{3}\right)\).

Using a calculator, \(\tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ\).

Since tangent has a period of \(180^\circ\), the general solution is:

\(2x + 15^\circ = 53.13^\circ + 180^\circ k\) or \(2x + 15^\circ = 233.13^\circ + 180^\circ k\), where \(k\) is an integer.

For \(k = 0\), solve for \(x\):

\(2x + 15^\circ = 53.13^\circ\)

\(2x = 53.13^\circ - 15^\circ\)

\(2x = 38.13^\circ\)

\(x = 19.065^\circ\)

Rounding to one decimal place, \(x = 19.1^\circ\).

For the second solution:

\(2x + 15^\circ = 233.13^\circ\)

\(2x = 233.13^\circ - 15^\circ\)

\(2x = 218.13^\circ\)

\(x = 109.065^\circ\)

Rounding to one decimal place, \(x = 109.1^\circ\).

Thus, the solutions are \(x = 19.1^\circ\) or \(x = 109.1^\circ\).

Start with the equation \(\sin 2x + 3 \cos 2x = 0\).

Rearrange to find \(\tan 2x\):

\(\tan 2x = -3\).

Find the angle \(2x\) using the inverse tangent function:

\(2x = 180^\circ - 71.6^\circ\) or \(2x = 360^\circ - 71.6^\circ\).

This gives \(2x = 108.4^\circ\) or \(2x = 288.4^\circ\).

Divide by 2 to solve for \(x\):

\(x = 54.2^\circ\) or \(x = 144.2^\circ\).

(i) Start with the equation \(\sin \theta + \cos \theta = 2(\sin \theta - \cos \theta)\).

Let \(s = \sin \theta\) and \(c = \cos \theta\).

Then the equation becomes \(s + c = 2s - 2c\).

Rearrange to get \(s + c = 2s - 2c \Rightarrow s + c = 2s - 2c \Rightarrow s = 3c\).

Divide both sides by \(c\) (assuming \(c \neq 0\)) to get \(\frac{s}{c} = 3\).

Thus, \(\tan \theta = 3\).

(ii) From \(\tan \theta = 3\), find \(\theta\) in the range \(0^\circ \leq \theta \leq 360^\circ\).

The principal value is \(\theta = \tan^{-1}(3) \approx 71.6^\circ\).

The other solution in the range is \(\theta = 180^\circ + 71.6^\circ = 251.6^\circ\).

Thus, \(\theta = 71.6^\circ\) or \(251.6^\circ\).

(a) Start with \(\frac{\tan x + \sin x}{\tan x - \sin x} = k\).

Multiply numerator and denominator by \(\cos x\):

\(\frac{\sin x + \sin x \cos x}{\sin x - \sin x \cos x} = k\).

Rearrange to \(\frac{1 + \cos x}{1 - \cos x} = k\).

(b) From \(\frac{1 + \cos x}{1 - \cos x} = k\), cross-multiply to get:

\(k - k \cos x = 1 + \cos x\).

Rearrange to \(k - 1 = (k + 1) \cos x\).

Thus, \(\cos x = \frac{k - 1}{k + 1}\).

(c) Substitute \(k = 4\) into \(\cos x = \frac{k - 1}{k + 1}\):

\(\cos x = \frac{4 - 1}{4 + 1} = \frac{3}{5}\).

Find \(x\) such that \(\cos x = \frac{3}{5}\) within the range \(-\pi < x < \pi\).

The solutions are \(x = \pm 0.927\) (only solutions in the given range).

Given the equation \(\sin 3x + 2 \cos 3x = 0\).

We can rewrite this as \(\tan 3x = -2\) by dividing both sides by \(\cos 3x\).

Solving \(\tan 3x = -2\), we find the principal value:

\(3x = \tan^{-1}(-2) \approx -63.4^\circ\).

Since \(\tan\) is periodic with period \(180^\circ\), the general solution for \(3x\) is:

\(3x = -63.4^\circ + 180^\circ n\), where \(n\) is an integer.

Solving for \(x\), we have:

\(x = \frac{-63.4^\circ + 180^\circ n}{3}\).

We need \(0^\circ \leq x \leq 180^\circ\), so:

\(0^\circ \leq \frac{-63.4^\circ + 180^\circ n}{3} \leq 180^\circ\).

Solving for \(n\), we find the valid values:

For \(n = 1\), \(x = \frac{116.6^\circ}{3} = 38.9^\circ\).

For \(n = 2\), \(x = \frac{296.6^\circ}{3} = 98.9^\circ\).

For \(n = 3\), \(x = \frac{476.6^\circ}{3} = 158.9^\circ\).

Thus, the solutions are \(x = 38.9^\circ, 98.9^\circ, 158.9^\circ\).