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: \[ \boxed{1 + \sin^2 \theta = 3 \cos^2 \theta}. \]
The number of solutions of the equation $ \cos 2\theta \cos \left( \frac{\theta}{2} \right) + \cos \left( \frac{5\theta}{2} \right) = 2 \cos^3 \left( \frac{5\theta}{2} \right) $ in the interval \(\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\) is:
Observe the following data given in the table. (\(K_H\) = Henry's law constant)
| Gas | CO₂ | Ar | HCHO | CH₄ |
|---|---|---|---|---|
| \(K_H\) (k bar at 298 K) | 1.67 | 40.3 | \(1.83 \times 10^{-5}\) | 0.413 |
The correct order of their solubility in water is
For a first order decomposition of a certain reaction, rate constant is given by the equation
\(\log k(s⁻¹) = 7.14 - \frac{1 \times 10^4 K}{T}\). The activation energy of the reaction (in kJ mol⁻¹) is (\(R = 8.3 J K⁻¹ mol⁻¹\))
Note: The provided value for R is 8.3. We will use the more precise value R=8.314 J K⁻¹ mol⁻¹ for accuracy, as is standard.