Consider f, g and h be three real valued function defined on R. Let `f(x)=sin3x+cosx,g(x)=cos3x+sinx` and `h(x)=f^(2)(x)+g^(2)(x).` h(x) = 4

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to find the values of \( x \) such that \( h(x) = 4 \), where \( h(x) = f^2(x) + g^2(x) \), \( f(x) = \sin(3x) + \cos(x) \), and \( g(x) = \cos(3x) + \sin(x) \).

### Step 1: Write down the functions
We have:
- \( f(x) = \sin(3x) + \cos(x) \)
- \( g(x) = \cos(3x) + \sin(x) \)

### Step 2: Express \( h(x) \)
Now, we can express \( h(x) \):
\[ h(x) = f^2(x) + g^2(x) = (\sin(3x) + \cos(x))^2 + (\cos(3x) + \sin(x))^2 \]

### Step 3: Expand \( f^2(x) \) and \( g^2(x) \)
Expanding both squares:
\[ f^2(x) = \sin^2(3x) + 2\sin(3x)\cos(x) + \cos^2(x) \]
\[ g^2(x) = \cos^2(3x) + 2\cos(3x)\sin(x) + \sin^2(x) \]

### Step 4: Combine the expansions
Now, combine \( f^2(x) \) and \( g^2(x) \):
\[ h(x) = (\sin^2(3x) + \cos^2(3x)) + (\cos^2(x) + \sin^2(x)) + 2(\sin(3x)\cos(x) + \cos(3x)\sin(x)) \]

### Step 5: Use the Pythagorean identity
Using the identity \( \sin^2(a) + \cos^2(a) = 1 \):
\[ h(x) = 1 + 1 + 2(\sin(3x)\cos(x) + \cos(3x)\sin(x)) \]
\[ h(x) = 2 + 2\sin(3x + x) \]
\[ h(x) = 2 + 2\sin(4x) \]

### Step 6: Set \( h(x) = 4 \)
Now, we set \( h(x) = 4 \):
\[ 2 + 2\sin(4x) = 4 \]

### Step 7: Solve for \( \sin(4x) \)
Subtract 2 from both sides:
\[ 2\sin(4x) = 2 \]
Divide by 2:
\[ \sin(4x) = 1 \]

### Step 8: Find general solutions for \( 4x \)
The general solution for \( \sin(4x) = 1 \) is:
\[ 4x = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]

### Step 9: Solve for \( x \)
Now, divide by 4:
\[ x = \frac{\pi}{8} + \frac{n\pi}{2} \quad (n \in \mathbb{Z}) \]

### Final Answer
Thus, the solution for \( x \) is:
\[ x = \frac{\pi}{8} + \frac{n\pi}{2} \quad (n \in \mathbb{Z}) \]