If `sin^6 theta = 1+ cos^4 3 theta` then the most general value of `theta `is

Options

A. `(2n + 1/2)pi`
B. `(2n + 1)pi/6`
C. `(n + 1/2) pi/2`
D. none of these

Answer

To solve the equation \( \sin^6 \theta = 1 + \cos^4 3\theta \), we will follow these steps:

### Step 1: Analyze the Equation
We start with the equation:
\[ \sin^6 \theta = 1 + \cos^4 3\theta \]
We know that \( \sin^6 \theta \) is always less than or equal to 1 since the maximum value of \( \sin \theta \) is 1. Therefore, the left-hand side (LHS) is always less than or equal to 1.

### Step 2: Understand the Right Side
The right-hand side (RHS) is \( 1 + \cos^4 3\theta \). Since \( \cos^4 3\theta \) is a non-negative quantity (as any power of cosine is non-negative), the RHS is always greater than or equal to 1.

### Step 3: Set the Equation for Equality
For the two sides to be equal, we must have:
\[ \sin^6 \theta = 1 \]
This implies:
\[ \sin^2 \theta = 1 \]
Taking the square root gives:
\[ \sin \theta = \pm 1 \]

### Step 4: Solve for \( \theta \)
The solutions for \( \sin \theta = 1 \) and \( \sin \theta = -1 \) are:
1. \( \sin \theta = 1 \) gives \( \theta = \frac{\pi}{2} + 2n\pi \) for \( n \in \mathbb{Z} \).
2. \( \sin \theta = -1 \) gives \( \theta = \frac{3\pi}{2} + 2n\pi \) for \( n \in \mathbb{Z} \).

### Step 5: Combine the Solutions
Thus, the general solutions can be combined:
\[ \theta = \frac{\pi}{2} + 2n\pi \quad \text{and} \quad \theta = \frac{3\pi}{2} + 2n\pi \]
We can express this as:
\[ \theta = \frac{\pi}{2} + n\pi \quad \text{for } n \in \mathbb{Z} \]

### Final Answer
The most general value of \( \theta \) is:
\[ \theta = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \]