The number of solutions of the equation ` cos 4x+6=7 cos 2x ` , when ` x in[315^(@),317^(@)]` is

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the equation \( \cos 4x + 6 = 7 \cos 2x \) for \( x \) in the interval \([315^\circ, 317^\circ]\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos 4x + 6 = 7 \cos 2x \]
We can rearrange this to:
\[ \cos 4x = 7 \cos 2x - 6 \]

### Step 2: Use the double angle identity
We know that:
\[ \cos 4x = 2 \cos^2 2x - 1 \]
Substituting this into our equation gives:
\[ 2 \cos^2 2x - 1 = 7 \cos 2x - 6 \]

### Step 3: Rearrange the equation
Now, rearranging the equation, we have:
\[ 2 \cos^2 2x - 7 \cos 2x + 5 = 0 \]

### Step 4: Substitute \( t = \cos 2x \)
Let \( t = \cos 2x \). Then the equation becomes:
\[ 2t^2 - 7t + 5 = 0 \]

### Step 5: Factor the quadratic equation
We can factor this quadratic equation:
\[ (2t - 5)(t - 1) = 0 \]
This gives us the solutions:
\[ t = \frac{5}{2} \quad \text{and} \quad t = 1 \]

### Step 6: Analyze the solutions
Since \( t = \cos 2x \), we need to check if these values are valid:
1. \( t = \frac{5}{2} \) is not valid since the range of cosine is \([-1, 1]\).
2. \( t = 1 \) is valid.

### Step 7: Solve for \( x \)
For \( t = 1 \):
\[ \cos 2x = 1 \]
This implies:
\[ 2x = 0^\circ + 360^\circ n \quad \text{for } n \in \mathbb{Z} \]
Thus:
\[ x = 0^\circ + 180^\circ n \]

### Step 8: Find solutions in the given interval
Now we need to find \( x \) in the interval \([315^\circ, 317^\circ]\):
- The possible values of \( x \) are \( 0^\circ, 180^\circ, 360^\circ, \ldots \)
- None of these values fall within the interval \([315^\circ, 317^\circ]\).

### Conclusion
Thus, there are no solutions for the equation \( \cos 4x + 6 = 7 \cos 2x \) in the interval \([315^\circ, 317^\circ]\).