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

Answer 2

Step 1: Given equation and range
We are given that θ ∈ [0, π/4] and it satisfies \[ 4\cos\theta - 3\sin\theta = 1. \] We must find the value of \(\cos\theta.\)

Step 2: Express in a single cosine form
We can combine 4cosθ − 3sinθ in the form \( R\cos(\theta + \alpha) \).
Compute \(R = \sqrt{4^2 + (-3)^2} = 5.\)
Let \(\tan\alpha = \frac{3}{4} \Rightarrow \alpha = \tan^{-1}(3/4).\)
Then \[ 4\cos\theta - 3\sin\theta = 5\cos(\theta + \alpha). \] So the equation becomes \[ 5\cos(\theta + \alpha) = 1 \Rightarrow \cos(\theta + \alpha) = \frac{1}{5}. \] Hence \[ \theta + \alpha = \cos^{-1}\!\left(\frac{1}{5}\right). \] Therefore, \[ \theta = \cos^{-1}\!\left(\frac{1}{5}\right) - \alpha. \] Since α is small (tan⁻¹(3/4) ≈ 36.87°) and \(\cos^{-1}(1/5)\) ≈ 78.46°, θ ≈ 41.6° ≈ 0.726 rad, which lies in [0, π/4] only approximately if consistent with the problem’s constraints (here it’s indeed a valid positive acute value).

Step 3: Find sinθ and cosθ using the given linear equation and sin² + cos² = 1
Let \(\cos\theta = c,\ \sin\theta = s.\) Then \[ 4c - 3s = 1, \quad s^2 + c^2 = 1. \] From the first, \(s = \frac{4c - 1}{3}.\) Substitute into the identity: \[ \left(\frac{4c - 1}{3}\right)^2 + c^2 = 1. \] Simplify: \[ \frac{(4c - 1)^2}{9} + c^2 = 1 \Rightarrow (16c^2 - 8c + 1) + 9c^2 = 9 \Rightarrow 25c^2 - 8c - 8 = 0. \] Thus, \[ c = \frac{8 \pm \sqrt{64 + 800}}{50} = \frac{8 \pm \sqrt{864}}{50} = \frac{8 \pm 12\sqrt{6}}{50}. \] Since θ ∈ [0, π/4], both sinθ and cosθ are positive, so we take the positive root: \[ \cos\theta = \frac{8 + 12\sqrt{6}}{50}. \] Simplify the fraction by dividing numerator and denominator by 2: \[ \cos\theta = \frac{4 + 6\sqrt{6}}{25}. \] Now rationalize this to match the required form: \[ \frac{4 + 6\sqrt{6}}{25} = \frac{4}{3\sqrt{6} - 2}. \] Hence, \[ \cos\theta = \frac{4}{3\sqrt{6} - 2}. \]

Final answer

\(\frac{4}{3\sqrt{6} - 2}\)