Question:
If \(x, y \in [0, 2\pi]\) then the total number of ordered pairs \((x, y)\) satisfying \(\sin x \cdot \cos y = 1\) is equal to
If \(x, y \in [0, 2\pi]\) then the total number of ordered pairs \((x, y)\) satisfying \(\sin x \cdot \cos y = 1\) is equal to
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For product to be 1, both factors must be 1 or both -1.
Updated On: Apr 23, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Formula / Definition}
\[ \sin x \cdot \cos y = 1 \]
Step 2: Calculation / Simplification}
Since \(|\sin x| \leq 1\) and \(|\cos y| \leq 1\), product = 1 only if:
Case 1: \(\sin x = 1, \cos y = 1\)
\(x = \pi/2\); \(y = 0, 2\pi\) \(\Rightarrow (\pi/2, 0), (\pi/2, 2\pi)\)
Case 2: \(\sin x = -1, \cos y = -1\)
\(x = 3\pi/2\); \(y = \pi\) \(\Rightarrow (3\pi/2, \pi)\)
Total = 3 ordered pairs.
Step 3: Final Answer
\[ 3 \]
\[ \sin x \cdot \cos y = 1 \]
Step 2: Calculation / Simplification}
Since \(|\sin x| \leq 1\) and \(|\cos y| \leq 1\), product = 1 only if:
Case 1: \(\sin x = 1, \cos y = 1\)
\(x = \pi/2\); \(y = 0, 2\pi\) \(\Rightarrow (\pi/2, 0), (\pi/2, 2\pi)\)
Case 2: \(\sin x = -1, \cos y = -1\)
\(x = 3\pi/2\); \(y = \pi\) \(\Rightarrow (3\pi/2, \pi)\)
Total = 3 ordered pairs.
Step 3: Final Answer
\[ 3 \]
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