Question:
The number of integral values of \(n\), for which the equation
\[
3\cos x + 5\sin x = 2n + 1
\]
has a solution, is:
The number of integral values of \(n\), for which the equation
\[
3\cos x + 5\sin x = 2n + 1
\]
has a solution, is:
Show 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.
Updated On: Feb 9, 2026
- \(4\)
- \(6\)
- \(8\)
- \(10\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
An expression of the form \(a\cos x + b\sin x\) can be written as \[ \sqrt{a^2+b^2}\sin(x+\phi) \]
Hence, its range is \[ -\sqrt{a^2+b^2} \le a\cos x + b\sin x \le \sqrt{a^2+b^2}. \]
Step 1: Find the range of the left-hand side \[ 3\cos x + 5\sin x \] Its maximum value is: \[ \sqrt{3^2+5^2}=\sqrt{34} \] Hence, \[ -\sqrt{34} \le 3\cos x + 5\sin x \le \sqrt{34} \]
Step 2: Apply the condition for solution For the equation to have a solution, \[ -\sqrt{34} \le 2n+1 \le \sqrt{34} \]
Step 3: Solve the inequality \[ -\sqrt{34}-1 \le 2n \le \sqrt{34}-1 \] \[ -3.415<n<2.415 \]
Step 4: Count integral values of \(n\) The integers satisfying the inequality are: \[ n=-3,-2,-1,0,1,2 \] Total number of integral values: \[ 6 \]
An expression of the form \(a\cos x + b\sin x\) can be written as \[ \sqrt{a^2+b^2}\sin(x+\phi) \]
Hence, its range is \[ -\sqrt{a^2+b^2} \le a\cos x + b\sin x \le \sqrt{a^2+b^2}. \]
Step 1: Find the range of the left-hand side \[ 3\cos x + 5\sin x \] Its maximum value is: \[ \sqrt{3^2+5^2}=\sqrt{34} \] Hence, \[ -\sqrt{34} \le 3\cos x + 5\sin x \le \sqrt{34} \]
Step 2: Apply the condition for solution For the equation to have a solution, \[ -\sqrt{34} \le 2n+1 \le \sqrt{34} \]
Step 3: Solve the inequality \[ -\sqrt{34}-1 \le 2n \le \sqrt{34}-1 \] \[ -3.415<n<2.415 \]
Step 4: Count integral values of \(n\) The integers satisfying the inequality are: \[ n=-3,-2,-1,0,1,2 \] Total number of integral values: \[ 6 \]
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