The equation `k cos x-3sinx=k+1` is solvable only if

Options

A. `k in (-oo, 4)`
B. `k in (-oo, 4]`
C. `k in (4, oo)`
D. `k in [4, oo)`

Answer

To solve the equation \( k \cos x - 3 \sin x = k + 1 \) for the values of \( k \) for which it is solvable, we can follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ k \cos x - 3 \sin x = k + 1 \]
Rearranging gives:
\[ k \cos x - 3 \sin x - k - 1 = 0 \]

### Step 2: Identify the coefficients
This can be treated as a linear combination of \( \cos x \) and \( \sin x \):
\[ k \cos x - 3 \sin x = k + 1 \]
Let \( A = k \) and \( B = -3 \). The equation can be expressed as:
\[ A \cos x + B \sin x = k + 1 \]

### Step 3: Use the Ranges of Trigonometric Functions
The maximum value of \( A \cos x + B \sin x \) is given by:
\[ \sqrt{A^2 + B^2} \]
Thus, we need:
\[ |k + 1| \leq \sqrt{k^2 + (-3)^2} \]
This simplifies to:
\[ |k + 1| \leq \sqrt{k^2 + 9} \]

### Step 4: Square both sides
To eliminate the absolute value, we consider two cases:
1. \( k + 1 \geq 0 \)
2. \( k + 1 < 0 \)

Squaring both sides gives:
\[ (k + 1)^2 \leq k^2 + 9 \]

### Step 5: Expand and simplify
Expanding the left side:
\[ k^2 + 2k + 1 \leq k^2 + 9 \]
Subtract \( k^2 \) from both sides:
\[ 2k + 1 \leq 9 \]
This simplifies to:
\[ 2k \leq 8 \]
Thus:
\[ k \leq 4 \]

### Step 6: Consider the second case
For the second case, if \( k + 1 < 0 \), we have:
\[ -(k + 1) \leq \sqrt{k^2 + 9} \]
Squaring both sides gives:
\[ (k + 1)^2 \leq k^2 + 9 \]
This leads to the same inequality:
\[ 2k + 1 \leq 9 \]
Thus, we again find:
\[ k \leq 4 \]

### Conclusion
The equation \( k \cos x - 3 \sin x = k + 1 \) is solvable only if:
\[ k \leq 4 \]