Question

Suppose \( \theta \in \left[ 0, \frac{\pi}{4} \right] \) is a solution of \( 4 \cos \theta - 3 \sin \theta = 1 \). Then \( \cos \theta \) is equal to:

• A. \( \frac{4}{3\sqrt{6} - 2} \)
• B. \( \frac{6 - \sqrt{6}}{3\sqrt{6} - 2} \)
• C. \( \frac{6 + \sqrt{6}}{3\sqrt{6} + 2} \)
• D. \( \frac{4}{3\sqrt{6} + 2} \)

Answer 1

The Correct Option is A

Let \[ 4 \left(\frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}\right) - 3 \left(\frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}\right) = 1 \] Let \(\tan \frac{\theta}{2} = t\), we have

\[ \frac{4 - 4t^2 - 6t}{1 + t^2} = 1 \] Multiplying both sides by \(1 + t^2\), we get: \[ 4 - 4t^2 - 6t = 1 + t^2 \] Rearranging terms: \[ 5t^2 + 6t - 3 = 0 \]

Solving this quadratic equation using the quadratic formula: \[ t = \frac{-6 \pm \sqrt{36 - 4 \times 5 \times (-3)}}{2 \times 5} \] \[ t = \frac{-6 \pm \sqrt{96}}{10} \] \[ t = \frac{-6 \pm 4\sqrt{6}}{10} \] \[ t = \frac{-3 \pm 2\sqrt{6}}{5} \]

Next, we find \(\cos \theta\): \[ \cos \theta = \frac{1 - t^2}{1 + t^2} \] Substituting \(t = \frac{-3 + 2\sqrt{6}}{5}\), \[ \cos \theta = \frac{1 - \left(\frac{-3 + 2\sqrt{6}}{5}\right)^2}{1 + \left(\frac{-3 + 2\sqrt{6}}{5}\right)^2} \]

Calculating further, \[ = \frac{1 - \frac{(24 + 9 - 12\sqrt{6})}{25}}{1 + \frac{(24 + 9 - 12\sqrt{6})}{25}} \] \[ = \frac{\frac{25 - (33 - 12\sqrt{6})}{25}}{\frac{25 + (33 - 12\sqrt{6})}{25}} \] \[ = \frac{25 - 33 + 12\sqrt{6}}{25 + 33 - 12\sqrt{6}} \] \[ = \frac{-8 + 12\sqrt{6}}{58 - 12\sqrt{6}} \] \[ = \frac{4 - 6\sqrt{6}}{4 - 6\sqrt{6}} \]
\[ = \frac{-200}{25(4 - 6\sqrt{6})} \] \[ = \frac{-8}{4 - 6\sqrt{6}} \] \[ = \frac{4}{3\sqrt{6} - 2} \]