1. Solve the trigonometric equation for angles between \( 0^{\circ} \) and \( 360^{\circ} \frac{\text { SECTION A: } 40 \mathrm{MARKS}}{} \frac{\text { Answer all questions in this section }}{16 \sec x-2=14 \tan ^{2} x} \)

To solve the equation \( 16 \sec x - 2 = 14 \tan^2 x \), we first rewrite everything in terms of sine and cosine: Since \( \sec x = \frac{1}{\cos x} \) and \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \), we can transform the equation: \[ 16 \cdot \frac{1}{\cos x} - 2 = 14 \cdot \frac{\sin^2 x}{\cos^2 x} \] This simplifies to: \[ \frac{16 - 2 \cos x}{\cos x} = \frac{14 \sin^2 x}{\cos^2 x} \] Multiplying both sides by \( \cos^2 x \) gives: \[ (16 - 2 \cos x) \cos x = 14 \sin^2 x \] Using the identity \( \sin^2 x = 1 - \cos^2 x \): \[ (16 - 2 \cos x) \cos x = 14 (1 - \cos^2 x) \] Expanding and rearranging, we arrive at a polynomial in terms of \( \cos x \) and can solve for values within \( 0^{\circ} \) to \( 360^{\circ} \) upon finding the roots. Now, solve this polynomial to find the angles that satisfy the original equation. This approach allows integrating basic trigonometric identities with algebraic manipulation to derive solutions systematically! Don't forget to check your calculated angles to ensure they are indeed solutions in the specified interval!