The value of `k` if the equation `2cosx+cos2k x=3` has only one solution is 2 (b) 2 (c) `sqrt(2)` (d) `1/2`

Options

A. 0
B. 2
C. `sqrt(2)`
D. `1//2`

Explanation

To find the value of \( k \) such that the equation \( 2 \cos x + \cos(2kx) = 3 \) has only one solution, we can follow these steps:

### Step 1: Analyze the Equation
The given equation is:
\[ 2 \cos x + \cos(2kx) = 3 \]
Since the maximum value of \( \cos x \) is 1, the maximum value of \( 2 \cos x \) is \( 2 \). Therefore, for the left-hand side to equal 3, we must have:
\[ 2 \cos x + \cos(2kx) = 3 \implies \cos(2kx) = 3 - 2 \cos x \]
To satisfy this equation, \( 3 - 2 \cos x \) must also be in the range of the cosine function, which is \([-1, 1]\).

### Step 2: Determine the Range
From the equation \( 3 - 2 \cos x \):
- The minimum value occurs when \( \cos x = 1 \):
\[ 3 - 2(1) = 1 \]
- The maximum value occurs when \( \cos x = -1 \):
\[ 3 - 2(-1) = 5 \]
Thus, for \( 3 - 2 \cos x \) to be in the range of \([-1, 1]\), we have:
\[ 1 \leq 3 - 2 \cos x \leq 1 \]
This implies that:
\[ 3 - 2 \cos x = 1 \implies 2 \cos x = 2 \implies \cos x = 1 \]
This means \( x = 2n\pi \) for \( n \in \mathbb{Z} \).

### Step 3: Solve for \( k \)
Now substituting \( \cos x = 1 \) into the equation:
\[ \cos(2kx) = 3 - 2(1) = 1 \]
This implies:
\[ \cos(2kx) = 1 \implies 2kx = 2m\pi \quad (m \in \mathbb{Z}) \]
Thus,
\[ kx = m\pi \]
For \( x = 2n\pi \):
\[ k(2n\pi) = m\pi \implies k = \frac{m}{2n} \]
To have only one solution for \( x \), \( k \) must be such that the ratio \( \frac{m}{2n} \) is unique. This occurs when \( k \) is an irrational number.

### Step 4: Conclusion
Among the given options for \( k \):
- (a) 0
- (b) 2
- (c) \( \sqrt{2} \)
- (d) \( \frac{1}{2} \)

Only \( \sqrt{2} \) is irrational. Therefore, the value of \( k \) for which the equation has only one solution is:
\[ \boxed{\sqrt{2}} \]