The general solution of the equation ` sin^(2)x + cos^(2)3x =1 ` is equal to :
(where ` n in I `)

Options

A. ` x = (n pi)/(2) `
B. ` x = n pi + (pi)/(4) `
C. ` x= (n pi)/(4) `
D. ` x = n pi + (pi)/(2) `

Answer

To solve the equation \( \sin^2 x + \cos^2 3x = 1 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin^2 x + \cos^2 3x = 1 \]
Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can express \( \cos^2 3x \) in terms of \( \sin^2 x \):
\[ \cos^2 3x = 1 - \sin^2 3x \]
So, we rewrite the equation as:
\[ \sin^2 x + (1 - \sin^2 3x) = 1 \]
This simplifies to:
\[ \sin^2 x - \sin^2 3x = 0 \]

### Step 2: Factor the equation
We can factor the equation:
\[ \sin^2 x = \sin^2 3x \]
This implies:
\[ \sin^2 x - \sin^2 3x = 0 \]
Using the identity \( a^2 - b^2 = (a - b)(a + b) \), we can write:
\[ (\sin x - \sin 3x)(\sin x + \sin 3x) = 0 \]

### Step 3: Solve each factor
We now have two cases to consider:

**Case 1:**
\[ \sin x - \sin 3x = 0 \]
This implies:
\[ \sin x = \sin 3x \]
The general solution for \( \sin A = \sin B \) is:
\[ x = 3x + 2n\pi \quad \text{or} \quad x = \pi - 3x + 2n\pi \]
From the first equation:
\[ -2x = 2n\pi \implies x = -n\pi \]
From the second equation:
\[ 4x = \pi + 2n\pi \implies x = \frac{\pi}{4} + \frac{n\pi}{2} \]

**Case 2:**
\[ \sin x + \sin 3x = 0 \]
This implies:
\[ \sin x = -\sin 3x \]
The general solution for \( \sin A = -\sin B \) is:
\[ x = -3x + 2n\pi \quad \text{or} \quad x = \pi + 3x + 2n\pi \]
From the first equation:
\[ 4x = 2n\pi \implies x = \frac{n\pi}{2} \]
From the second equation:
\[ -2x = \pi + 2n\pi \implies x = -\frac{\pi}{2} - n\pi \]

### Step 4: Combine the solutions
Now we combine all the solutions we have found:
1. From Case 1: \( x = -n\pi \) and \( x = \frac{\pi}{4} + \frac{n\pi}{2} \)
2. From Case 2: \( x = \frac{n\pi}{2} \) and \( x = -\frac{\pi}{2} - n\pi \)

### Step 5: Identify the general solution
The most comprehensive solution that includes all cases is:
\[ x = \frac{n\pi}{4} \]
where \( n \in \mathbb{Z} \).

Thus, the general solution of the equation \( \sin^2 x + \cos^2 3x = 1 \) is:
\[ \boxed{x = n\frac{\pi}{4}} \]