To express \(\sec \theta = 3 \cos \theta + \tan \theta\) as a quadratic in \(\sin \theta\), we start by using trigonometric identities:

\(\sec \theta = \frac{1}{\cos \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).

Substitute these into the equation:

\(\frac{1}{\cos \theta} = 3 \cos \theta + \frac{\sin \theta}{\cos \theta}\)

Multiply through by \(\cos \theta\) to clear the fractions:

\(1 = 3 \cos^2 \theta + \sin \theta\)

Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):

\(1 = 3(1 - \sin^2 \theta) + \sin \theta\)

Expand and rearrange:

\(1 = 3 - 3 \sin^2 \theta + \sin \theta\)

\(3 \sin^2 \theta - \sin \theta - 2 = 0\)

This is a quadratic equation in \(\sin \theta\). Let \(x = \sin \theta\), then:

\(3x^2 - x - 2 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = -1\), \(c = -2\):

\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3}\)

\(x = \frac{1 \pm \sqrt{1 + 24}}{6}\)

\(x = \frac{1 \pm 5}{6}\)

\(x = 1 \text{ or } x = -\frac{2}{3}\)

Since \(x = \sin \theta\), and \(-1 \leq \sin \theta \leq 1\), only \(x = -\frac{2}{3}\) is valid in the interval \(-90^\circ < \theta < 90^\circ\).

Thus, \(\sin \theta = -\frac{2}{3}\).

Find \(\theta\):

\(\theta = \arcsin\left(-\frac{2}{3}\right) \approx -41.8^\circ\)

(i) To prove the identity \(\cos 4\theta - 4\cos 2\theta \equiv 8\sin^4\theta - 3\):

Use double angle formulas:

\(\cos 4\theta = 2\cos^2 2\theta - 1\)

\(\cos 2\theta = 2\cos^2 \theta - 1\)

Substitute \(\cos 2\theta = 1 - 2\sin^2 \theta\):

\(\cos 4\theta = 2(1 - 2\sin^2 \theta)^2 - 1\)

\(= 2(1 - 4\sin^2 \theta + 4\sin^4 \theta) - 1\)

\(= 2 - 8\sin^2 \theta + 8\sin^4 \theta - 1\)

\(= 8\sin^4 \theta - 8\sin^2 \theta + 1\)

\(\cos 4\theta - 4\cos 2\theta = 8\sin^4 \theta - 8\sin^2 \theta + 1 - 4(1 - 2\sin^2 \theta)\)

\(= 8\sin^4 \theta - 8\sin^2 \theta + 1 - 4 + 8\sin^2 \theta\)

\(= 8\sin^4 \theta - 3\)

Thus, the identity is proven.

(ii) To solve \(\cos 4\theta = 4\cos 2\theta + 3\):

Using the identity \(\cos 4\theta - 4\cos 2\theta = 8\sin^4 \theta - 3\), set \(8\sin^4 \theta - 3 = 3\).

\(8\sin^4 \theta = 6\)

\(\sin^4 \theta = \frac{3}{4}\)

\(\sin^2 \theta = \frac{\sqrt{3}}{2}\)

\(\sin \theta = \pm \frac{\sqrt{3}}{2}\)

\(\theta = 60^\circ, 120^\circ, 240^\circ, 300^\circ\)

Adjust for the equation \(\cos 4\theta = 4\cos 2\theta + 3\):

\(\theta = 68.5^\circ, 111.5^\circ, 248.5^\circ, 291.5^\circ\)

1. Start with the given equation:

\(\csc \theta = 3 \sin \theta + \cot \theta\)

2. Use the identities \(\csc \theta = \frac{1}{\sin \theta}\) and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\):

\(\frac{1}{\sin \theta} = 3 \sin \theta + \frac{\cos \theta}{\sin \theta}\)

3. Multiply through by \(\sin \theta\) to clear the fractions:

\(1 = 3 \sin^2 \theta + \cos \theta\)

4. Use the Pythagorean identity \(\sin^2 \theta = 1 - \cos^2 \theta\):

\(1 = 3(1 - \cos^2 \theta) + \cos \theta\)

5. Simplify and rearrange to form a quadratic equation:

\(1 = 3 - 3\cos^2 \theta + \cos \theta\)

\(0 = 3\cos^2 \theta - \cos \theta - 2\)

6. Solve the quadratic equation \(3\cos^2 \theta - \cos \theta - 2 = 0\) using the quadratic formula \(\cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -1\), \(c = -2\):

\(\cos \theta = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 3 \times (-2)}}{2 \times 3}\)

\(\cos \theta = \frac{1 \pm \sqrt{1 + 24}}{6}\)

\(\cos \theta = \frac{1 \pm 5}{6}\)

\(\cos \theta = 1 \text{ or } \cos \theta = -\frac{2}{3}\)

7. Since \(\cos \theta = 1\) is not in the interval \(0^\circ < \theta < 180^\circ\), we use \(\cos \theta = -\frac{2}{3}\):

\(\theta = \cos^{-1}\left(-\frac{2}{3}\right) \approx 131.8^\circ\)

1. Use the identity \(\cot 2x = \frac{\cot^2 x - 1}{2\cot x}\) to express \(\cot 2x\) in terms of \(\cot x\).

2. Substitute into the equation: \(\frac{\cot^2 x - 1}{2\cot x} + \cot x = 3\).

3. Multiply through by \(2\cot x\) to clear the fraction: \(\cot^2 x - 1 + 2\cot^2 x = 6\cot x\).

4. Simplify to get: \(3\cot^2 x - 6\cot x - 1 = 0\).

5. Let \(y = \cot x\), then the equation becomes \(3y^2 - 6y - 1 = 0\).

6. Solve the quadratic equation using the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3, b = -6, c = -1\).

7. Calculate the discriminant: \(b^2 - 4ac = 36 + 12 = 48\).

8. Solve for \(y\): \(y = \frac{6 \pm \sqrt{48}}{6}\).

9. Simplify to find \(y = 3 \pm \sqrt{3}\).

10. Find \(x\) using \(x = \cot^{-1}(y)\).

11. Calculate \(x\) for \(y = 3 + \sqrt{3}\) and \(y = 3 - \sqrt{3}\) within the interval \(0^\circ < x < 180^\circ\).

12. The solutions are \(x = 24.9^\circ\) and \(x = 98.8^\circ\).

(i) We start with the expression \(\sin 2\alpha \sec \alpha\).

Using the identity \(\sin 2\alpha = 2 \sin \alpha \cos \alpha\), we have:

\(\sin 2\alpha \sec \alpha = 2 \sin \alpha \cos \alpha \cdot \sec \alpha\)

Since \(\sec \alpha = \frac{1}{\cos \alpha}\), substitute to get:

\(2 \sin \alpha \cos \alpha \cdot \frac{1}{\cos \alpha} = 2 \sin \alpha\)

Thus, the simplified expression is \(2 \sin \alpha\).

(ii) Given the equation \(3 \cos 2\beta + 7 \cos \beta = 0\), use the identity \(\cos 2\beta = 2 \cos^2 \beta - 1\):

Substitute to get:

\(3(2 \cos^2 \beta - 1) + 7 \cos \beta = 0\)

Simplify to form a quadratic equation:

\(6 \cos^2 \beta - 3 + 7 \cos \beta = 0\)

\(6 \cos^2 \beta + 7 \cos \beta - 3 = 0\)

Solving this quadratic equation using the quadratic formula \(\cos \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6\), \(b = 7\), \(c = -3\):

\(\cos \beta = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 6 \cdot (-3)}}{2 \cdot 6}\)

\(\cos \beta = \frac{-7 \pm \sqrt{49 + 72}}{12}\)

\(\cos \beta = \frac{-7 \pm \sqrt{121}}{12}\)

\(\cos \beta = \frac{-7 \pm 11}{12}\)

This gives two solutions:

\(\cos \beta = \frac{4}{12} = \frac{1}{3}\)

\(\cos \beta = \frac{-18}{12} = -\frac{3}{2}\)

Since \(\cos \beta\) must be within \([-1, 1]\), the valid solution is \(\cos \beta = \frac{1}{3}\).