If `|k|=5 and 0^(@) le theta le 360^(@) ` , then the number of different solution of ` 3 cos theta + 4 sin theta =k` is

Options

A. Zero
B. Two
C. One
D. Infinite

Answer

To solve the equation \(3 \cos \theta + 4 \sin \theta = k\) given that \(|k| = 5\) and \(0^\circ \leq \theta \leq 360^\circ\), we will follow these steps:

### Step 1: Determine the range of \(3 \cos \theta + 4 \sin \theta\)

The expression \(3 \cos \theta + 4 \sin \theta\) can be analyzed using the formula for the maximum and minimum values of \(a \cos \theta + b \sin \theta\), where \(a = 3\) and \(b = 4\).

The maximum value is given by:
\[ \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

The minimum value is:
\[ -\sqrt{a^2 + b^2} = -\sqrt{25} = -5 \]

### Step 2: Analyze the values of \(k\)

Since \(|k| = 5\), \(k\) can take two values: \(k = 5\) or \(k = -5\).

### Step 3: Solve for \(k = 5\)

For \(k = 5\):
\[ 3 \cos \theta + 4 \sin \theta = 5 \]
The maximum value of \(3 \cos \theta + 4 \sin \theta\) is 5, which means there exists at least one solution for this equation.

### Step 4: Solve for \(k = -5\)

For \(k = -5\):
\[ 3 \cos \theta + 4 \sin \theta = -5 \]
The minimum value of \(3 \cos \theta + 4 \sin \theta\) is -5, which means there also exists at least one solution for this equation.

### Step 5: Conclusion on the number of solutions

Since both \(k = 5\) and \(k = -5\) yield at least one solution, we need to determine how many solutions exist for each case.

The equation \(3 \cos \theta + 4 \sin \theta = k\) can be rewritten in the form \(R \sin(\theta + \phi) = k\), where \(R = 5\) and \(\phi\) is a phase shift. The equation \(R \sin(\theta + \phi) = k\) can have at most two solutions in the interval \(0^\circ \leq \theta < 360^\circ\) for each value of \(k\).

Thus, for \(k = 5\), there are 2 solutions, and for \(k = -5\), there are also 2 solutions.

### Final Answer

The total number of different solutions of \(3 \cos \theta + 4 \sin \theta = k\) is **4**.

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