Consider f, g and h be three real valued function defined on R.
Let `f(x)= sin 3x + cos x , g(x)= cos 3x + sin x ` and ` h(x) = f^(2)(x) + g^(2)(x) `
Q. General solution of the equation ` h(x) = 4 ` , is :
[where ` n in I ` ]

Options

A. ` (4n + 1) (pi)/(8) `
B. ` (8n + 1) (pi)/(8) `
C. ` (2n + 1) (pi)/(4) `
D. ` (7n + 1) (pi)/(4) `

Answer

To solve the problem, we need to find the general solution of the equation \( h(x) = 4 \), where:

- \( f(x) = \sin(3x) + \cos(x) \)
- \( g(x) = \cos(3x) + \sin(x) \)
- \( h(x) = f^2(x) + g^2(x) \)

### Step 1: Write down the expressions for \( f(x) \) and \( g(x) \)

We have:
\[ f(x) = \sin(3x) + \cos(x) \]
\[ g(x) = \cos(3x) + \sin(x) \]

### Step 2: Compute \( h(x) \)

We need to compute \( h(x) = f^2(x) + g^2(x) \).

Expanding \( f^2(x) \):
\[ f^2(x) = (\sin(3x) + \cos(x))^2 = \sin^2(3x) + \cos^2(x) + 2\sin(3x)\cos(x) \]

Expanding \( g^2(x) \):
\[ g^2(x) = (\cos(3x) + \sin(x))^2 = \cos^2(3x) + \sin^2(x) + 2\cos(3x)\sin(x) \]

### Step 3: Combine \( f^2(x) \) and \( g^2(x) \)

Now, we combine these results:
\[ h(x) = f^2(x) + g^2(x) = (\sin^2(3x) + \cos^2(x) + 2\sin(3x)\cos(x)) + (\cos^2(3x) + \sin^2(x) + 2\cos(3x)\sin(x)) \]

### Step 4: Use Pythagorean identity

Using the identity \( \sin^2(a) + \cos^2(a) = 1 \):
\[ \sin^2(3x) + \cos^2(3x) = 1 \]
\[ \sin^2(x) + \cos^2(x) = 1 \]

Thus, we can simplify \( h(x) \):
\[ h(x) = 1 + 1 + 2\sin(3x)\cos(x) + 2\cos(3x)\sin(x) \]
\[ h(x) = 2 + 2(\sin(3x)\cos(x) + \cos(3x)\sin(x)) \]

### Step 5: Use the sine addition formula

Using the sine addition formula, we have:
\[ \sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b) \]
So,
\[ \sin(3x + x) = \sin(4x) \]

Thus, we can rewrite \( h(x) \):
\[ h(x) = 2 + 2\sin(4x) \]

### Step 6: Set \( h(x) = 4 \)

Now we set this equal to 4:
\[ 2 + 2\sin(4x) = 4 \]
Subtracting 2 from both sides:
\[ 2\sin(4x) = 2 \]
Dividing by 2:
\[ \sin(4x) = 1 \]

### Step 7: Solve for \( x \)

The general solution for \( \sin(\theta) = 1 \) is:
\[ \theta = \frac{\pi}{2} + 2n\pi \quad (n \in \mathbb{Z}) \]
Thus, for our equation:
\[ 4x = \frac{\pi}{2} + 2n\pi \]
Dividing by 4:
\[ x = \frac{\pi}{8} + \frac{n\pi}{2} \quad (n \in \mathbb{Z}) \]

### Final Answer

The general solution of the equation \( h(x) = 4 \) is:
\[ x = \frac{\pi}{8} + \frac{n\pi}{2} \quad (n \in \mathbb{Z}) \]