The number of points in interval `[ - (pi)/(2) , (pi)/(2)], ` where the graphs of the curves ` y = cos x ` and ` y= sin 3x , -(pi)/(2) le x le (pi)/(2) ` intersects is

Answer

To find the number of points where the graphs of the curves \( y = \cos x \) and \( y = \sin 3x \) intersect in the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\), we need to solve the equation:

\[ \cos x = \sin 3x \]

### Step 1: Rewrite the equation
Using the identity \( \cos x = \sin\left(\frac{\pi}{2} - x\right) \), we can rewrite the equation as:

\[ \sin 3x - \cos x = 0 \]

### Step 2: Set up the equation
This can be rearranged to:

\[ \sin 3x - \sin\left(\frac{\pi}{2} - x\right) = 0 \]

### Step 3: Use the sine subtraction formula
Using the sine subtraction formula, we can express this as:

\[ 2 \cos\left(\frac{3x + \left(\frac{\pi}{2} - x\right)}{2}\right) \sin\left(\frac{3x - \left(\frac{\pi}{2} - x\right)}{2}\right) = 0 \]

This simplifies to:

\[ 2 \cos\left(\frac{3x + \frac{\pi}{2} - x}{2}\right) \sin\left(\frac{3x - \frac{\pi}{2} + x}{2}\right) = 0 \]

### Step 4: Solve the product equation
The product is zero if either factor is zero:

1. \( \cos\left(2x + \frac{\pi}{4}\right) = 0 \)
2. \( \sin(2x - \frac{\pi}{4}) = 0 \)

### Step 5: Solve \( \cos\left(2x + \frac{\pi}{4}\right) = 0 \)
The cosine function is zero at:

\[ 2x + \frac{\pi}{4} = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

Solving for \( x \):

\[ 2x = \frac{\pi}{4} + n\pi \implies x = \frac{\pi}{8} + \frac{n\pi}{2} \]

### Step 6: Determine valid \( n \) values
We need \( x \) in the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\):

- For \( n = -1 \): \( x = \frac{\pi}{8} - \frac{\pi}{2} = -\frac{3\pi}{8} \)
- For \( n = 0 \): \( x = \frac{\pi}{8} \)
- For \( n = 1 \): \( x = \frac{\pi}{8} + \frac{\pi}{2} = \frac{5\pi}{8} \) (not valid)

### Step 7: Solve \( \sin(2x - \frac{\pi}{4}) = 0 \)
The sine function is zero at:

\[ 2x - \frac{\pi}{4} = n\pi \quad (n \in \mathbb{Z}) \]

Solving for \( x \):

\[ 2x = \frac{\pi}{4} + n\pi \implies x = \frac{\pi}{8} + \frac{n\pi}{2} \]

### Step 8: Determine valid \( n \) values
Again, we need \( x \) in the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\):

- For \( n = -1 \): \( x = -\frac{3\pi}{8} \) (already counted)
- For \( n = 0 \): \( x = \frac{\pi}{8} \) (already counted)
- For \( n = 1 \): \( x = \frac{5\pi}{8} \) (not valid)

### Conclusion
The valid intersection points are:

1. \( x = -\frac{3\pi}{8} \)
2. \( x = \frac{\pi}{8} \)
3. \( x = \frac{\pi}{4} \)

Thus, the total number of intersection points is **3**.

### Final Answer:
The number of points of intersection in the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\) is **3**.

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