Question

The number of integral values of \(n\), for which the equation \[ 3\cos x + 5\sin x = 2n + 1 \] has a solution, is:

• A. \(4\)
• B. \(6\)
• C. \(8\)
• D. \(10\)

Hint:

For equations of the form \(a\cos x + b\sin x = k\), always compare \(k\) with \(\pm\sqrt{a^2+b^2}\) to determine solvability.

Answer 1

The Correct Option is B

We are given the equation:

\[ 3\cos x + 5\sin x = 2n + 1 \]

To find the number of integral values of \( n \) for which this equation has a solution, we first determine the range of the left-hand side.

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Step 1: Find the range of \(3\cos x + 5\sin x\)

For any expression of the form \(a\cos x + b\sin x\), the range is:

\[ [-\sqrt{a^2+b^2},\; \sqrt{a^2+b^2}] \]

Here, \(a = 3\) and \(b = 5\), so:

\[ \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \]

Hence, the range of \(3\cos x + 5\sin x\) is:

\[ -\sqrt{34} \le 3\cos x + 5\sin x \le \sqrt{34} \]

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Step 2: Apply the condition for solutions

For the equation to have a solution, the right-hand side must lie within this range:

\[ -\sqrt{34} \le 2n + 1 \le \sqrt{34} \]

Subtracting 1 throughout:

\[ -\sqrt{34} - 1 \le 2n \le \sqrt{34} - 1 \]

Dividing by 2:

\[ \frac{-\sqrt{34}-1}{2} \le n \le \frac{\sqrt{34}-1}{2} \]

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Step 3: Determine integer values

Since \(\sqrt{34} \approx 5.831\), the inequality becomes:

\[ -3.4155 \le n \le 2.4155 \]

The integer values of \(n\) in this range are:

\[ n = -3, -2, -1, 0, 1, 2 \]

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Final Answer

\[ \boxed{6} \]