If the equation `kcosx-3sinx=k+1` has a solution for `x` then

Options

A. `[4, oo)`
B. `[-4, 4]`
C. `(-oo, 4]`
D. none of these

Explanation

To solve the equation \( k \cos x - 3 \sin x = k + 1 \) for the condition that it has a solution for \( x \), we will follow these steps:

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

### Step 2: Squaring Both Sides
To eliminate the trigonometric functions, we can square both sides. However, it's more effective to express one side in terms of the other. Let's isolate \( 3 \sin x \):
\[ 3 \sin x = k \cos x - (k + 1) \]
Now, squaring both sides:
\[ (3 \sin x)^2 = (k \cos x - (k + 1))^2 \]
This simplifies to:
\[ 9 \sin^2 x = (k \cos x - k - 1)^2 \]

### Step 3: Expanding Both Sides
Expanding the right-hand side:
\[ 9 \sin^2 x = (k \cos x - k - 1)(k \cos x - k - 1) = k^2 \cos^2 x - 2(k + 1)k \cos x + (k + 1)^2 \]
Thus, we have:
\[ 9 \sin^2 x = k^2 \cos^2 x - 2k(k + 1) \cos x + (k + 1)^2 \]

### Step 4: Using the Pythagorean Identity
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can express \( \sin^2 x \) as \( 1 - \cos^2 x \):
\[ 9(1 - \cos^2 x) = k^2 \cos^2 x - 2k(k + 1) \cos x + (k + 1)^2 \]
This simplifies to:
\[ 9 - 9 \cos^2 x = k^2 \cos^2 x - 2k(k + 1) \cos x + (k + 1)^2 \]

### Step 5: Rearranging into a Quadratic Form
Rearranging gives us:
\[ (9 + k^2) \cos^2 x - 2k(k + 1) \cos x + (k + 1)^2 - 9 = 0 \]
This is a quadratic equation in \( \cos x \).

### Step 6: Determining Conditions for Solutions
For this quadratic equation to have real solutions, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Here, \( a = 9 + k^2 \), \( b = -2k(k + 1) \), and \( c = (k + 1)^2 - 9 \).

Calculating the discriminant:
\[ D = (-2k(k + 1))^2 - 4(9 + k^2)((k + 1)^2 - 9) \]
Setting this greater than or equal to zero will give us the condition for \( k \).

### Step 7: Solving the Inequality
After simplifying the discriminant condition, we find:
\[ 72k - 288 \leq 0 \]
This leads to:
\[ k \leq 4 \]

### Conclusion
Thus, the value of \( k \) must satisfy:
\[ k \in (-\infty, 4] \]