The number of solutions of the equation ` 1 +sin^(4) x = cos ^(2) 3x, x in [-(5pi)/(2),(5pi)/(2)]` is

Options

A. 3
B. 5
C. 7
D. 4

Answer

To find the number of solutions of the equation \( 1 + \sin^4 x = \cos^2 3x \) for \( x \) in the interval \([-5\pi/2, 5\pi/2]\), we will follow these steps:

### Step 1: Analyze the ranges of both sides of the equation
The left-hand side of the equation is \( 1 + \sin^4 x \). Since \( \sin^4 x \) ranges from \( 0 \) to \( 1 \), we can conclude that:
\[ 1 + \sin^4 x \text{ ranges from } 1 \text{ to } 2. \]

The right-hand side is \( \cos^2 3x \). The cosine function ranges from \( -1 \) to \( 1 \), so \( \cos^2 3x \) will range from \( 0 \) to \( 1 \).

### Step 2: Set up the equation based on the ranges
From the analysis, we see that the left-hand side \( 1 + \sin^4 x \) can only equal \( \cos^2 3x \) when both sides are equal to \( 1 \) (since \( \cos^2 3x \) cannot exceed \( 1 \)). Therefore, we set:
\[ 1 + \sin^4 x = 1 \implies \sin^4 x = 0. \]

### Step 3: Solve for \( \sin^4 x = 0 \)
The equation \( \sin^4 x = 0 \) implies:
\[ \sin x = 0. \]
The general solutions for \( \sin x = 0 \) are:
\[ x = n\pi, \quad n \in \mathbb{Z}. \]

### Step 4: Determine the values of \( n \) within the given interval
We need to find the integer values of \( n \) such that:
\[ -\frac{5\pi}{2} \leq n\pi \leq \frac{5\pi}{2}. \]
Dividing the entire inequality by \( \pi \):
\[ -\frac{5}{2} \leq n \leq \frac{5}{2}. \]
This implies:
\[ -2.5 \leq n \leq 2.5. \]
The integer values of \( n \) that satisfy this inequality are:
\[ n = -2, -1, 0, 1, 2. \]

### Step 5: Count the solutions
Counting these integer values, we find:
- \( n = -2 \) gives \( x = -2\pi \)
- \( n = -1 \) gives \( x = -\pi \)
- \( n = 0 \) gives \( x = 0 \)
- \( n = 1 \) gives \( x = \pi \)
- \( n = 2 \) gives \( x = 2\pi \)

Thus, the total number of solutions is \( 5 \).

### Final Answer
The number of solutions of the equation \( 1 + \sin^4 x = \cos^2 3x \) in the interval \([-5\pi/2, 5\pi/2]\) is \( \boxed{5} \).