The number of integral values of 'k' for which the equation 3sinx + 4 cosx = k + 1 has a solution, k `in` R is

Answer

To solve the equation \( 3 \sin x + 4 \cos x = k + 1 \) for integral values of \( k \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 3 \sin x + 4 \cos x = k + 1 \]
We can express \( 3 \sin x + 4 \cos x \) in the form \( R \sin(x + \alpha) \) where \( R = \sqrt{a^2 + b^2} \) and \( a = 3, b = 4 \).

### Step 2: Calculate \( R \)
Calculate \( R \):
\[ R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Find the range of \( 3 \sin x + 4 \cos x \)
The expression \( 3 \sin x + 4 \cos x \) can take values in the range \([-R, R]\):
\[ -5 \leq 3 \sin x + 4 \cos x \leq 5 \]

### Step 4: Set the boundaries for \( k + 1 \)
From the equation \( 3 \sin x + 4 \cos x = k + 1 \), we can set the boundaries:
\[ -5 \leq k + 1 \leq 5 \]

### Step 5: Solve for \( k \)
Now, we can solve for \( k \):
\[ -5 - 1 \leq k \leq 5 - 1 \]
\[ -6 \leq k \leq 4 \]

### Step 6: Determine the integral values of \( k \)
The integral values of \( k \) that satisfy this inequality are:
\[ k = -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4 \]

### Step 7: Count the integral values
Counting these values:
-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4 gives us a total of 11 integral values.

### Final Answer
The number of integral values of \( k \) for which the equation has a solution is:
\[ \boxed{11} \]