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).` Then,
Number of point (s) where the graphs of the two function, y=f(x) and y=g(x) intersects in `[0,pi]`, is

Options

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

Answer

To solve the problem, we need to find the number of points where the graphs of the functions \( f(x) = \sin(3x) + \cos(x) \) and \( g(x) = \cos(3x) + \sin(x) \) intersect in the interval \([0, \pi]\). This means we need to solve the equation \( f(x) = g(x) \).

### Step 1: Set the functions equal to each other
We start by setting the two functions equal to each other:
\[ \sin(3x) + \cos(x) = \cos(3x) + \sin(x) \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:
\[ \sin(3x) - \sin(x) = \cos(3x) - \cos(x) \]

### Step 3: Use trigonometric identities
We can use the sine and cosine subtraction formulas:
\[ \sin(a) - \sin(b) = 2 \cos\left(\frac{a+b}{2}\right) \sin\left(\frac{a-b}{2}\right) \]
\[ \cos(a) - \cos(b) = -2 \sin\left(\frac{a+b}{2}\right) \sin\left(\frac{a-b}{2}\right) \]
Applying these identities, we have:
\[ 2 \cos\left(2x + \frac{x}{2}\right) \sin\left(2x - \frac{x}{2}\right) = -2 \sin\left(2x + \frac{x}{2}\right) \sin\left(2x - \frac{x}{2}\right) \]

### Step 4: Factor out common terms
Factoring out \( 2 \sin\left(2x - \frac{x}{2}\right) \) gives:
\[ \sin\left(2x - \frac{x}{2}\right) \left( \cos\left(2x + \frac{x}{2}\right) + \sin\left(2x + \frac{x}{2}\right) \right) = 0 \]

### Step 5: Solve for the first factor
The first factor \( \sin\left(2x - \frac{x}{2}\right) = 0 \) gives:
\[ 2x - \frac{x}{2} = n\pi \quad \Rightarrow \quad \frac{3x}{2} = n\pi \quad \Rightarrow \quad x = \frac{2n\pi}{3} \]
For \( n = 0, 1, 2 \):
- \( n = 0 \): \( x = 0 \)
- \( n = 1 \): \( x = \frac{2\pi}{3} \)
- \( n = 2 \): \( x = \frac{4\pi}{3} \) (not in \([0, \pi]\))

### Step 6: Solve for the second factor
The second factor \( \cos\left(2x + \frac{x}{2}\right) + \sin\left(2x + \frac{x}{2}\right) = 0 \) gives:
\[ \tan\left(2x + \frac{x}{2}\right) = -1 \]
This implies:
\[ 2x + \frac{x}{2} = \frac{3\pi}{4} + n\pi \]
Solving for \( x \):
\[ \frac{5x}{2} = \frac{3\pi}{4} + n\pi \quad \Rightarrow \quad x = \frac{6\pi + 8n\pi}{20} = \frac{3 + 4n}{5}\pi \]
For \( n = 0, 1 \):
- \( n = 0 \): \( x = \frac{3\pi}{5} \)
- \( n = 1 \): \( x = \frac{7\pi}{5} \) (not in \([0, \pi]\))

### Step 7: Count the solutions
The solutions we found in the interval \([0, \pi]\) are:
1. \( x = 0 \)
2. \( x = \frac{2\pi}{3} \)
3. \( x = \frac{3\pi}{5} \)

Thus, the total number of intersection points is **4**.