We are provided with the equation: \[ 4 - 3 \cos \theta = 5 \sin \theta \cos \theta. \] Our objective is to determine the equation that satisfies the general solution.
Step 1: Rewriting the given equation: \[ 4 - 3 \cos \theta = 5 \sin \theta \cos \theta. \] Rearrange the terms to group those involving \( \cos \theta \): \[ 4 = 3 \cos \theta + 5 \sin \theta \cos \theta. \]
Step 2: Utilize the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \) to express the equation in terms of \( \sin 2\theta \). Since: \[ 5 \sin \theta \cos \theta = \frac{5}{2} \sin 2\theta, \] substituting this into the equation results in: \[ 4 = 3 \cos \theta + \frac{5}{2} \sin 2\theta. \] We begin by testing option (4), \( 1 + \sin^2 \theta = 3 \cos^2 \theta \), to check its validity.
Step 4: Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we rewrite \( \cos^2 \theta \) as: \[ \cos^2 \theta = 1 - \sin^2 \theta. \] Substituting this into the equation \( 1 + \sin^2 \theta = 3 \cos^2 \theta \): \[ 1 + \sin^2 \theta = 3(1 - \sin^2 \theta). \] Simplifying: \[ 1 + \sin^2 \theta = 3 - 3 \sin^2 \theta. \] Rearranging: \[ 1 + \sin^2 \theta + 3 \sin^2 \theta = 3, \] \[ 1 + 4 \sin^2 \theta = 3. \] Solving for \( \sin^2 \theta \): \[ 4 \sin^2 \theta = 2 \quad \Rightarrow \quad \sin^2 \theta = \frac{1}{2}. \] Since this equation holds, it confirms that option (4) is the correct answer. \bigskip Thus, the equation that satisfies the general solution is: \[ \boxed{1 + \sin^2 \theta = 3 \cos^2 \theta}. \]
If \( \sqrt{5} - i\sqrt{15} = r(\cos\theta + i\sin\theta), -\pi < \theta < \pi, \) then
\[ r^2(\sec\theta + 3\csc^2\theta) = \]
For \( n \in \mathbb{N} \), the largest positive integer that divides \( 81^n + 20n - 1 \) is \( k \). If \( S \) is the sum of all positive divisors of \( k \), then find \( S - k \).