<p>If <span>\(0 < x < 3\pi\)</span>, <span>\(0 < y < 3\pi\)</span> and <span>\(\cos x \cdot \sin y = 1\)</span>, then find the possible number of values of the ordered pair <span>\((x, y)\)</span>.</p>
Step-by-Step Solution
Key Concept: For a product of two bounded quantities to equal 1, both must achieve their extreme values simultaneously. Use the constraint that cosine and sine are bounded by [-1, 1].
<p><strong>Step 1:</strong> Since <span>\(\cos x \cdot \sin y = 1\)</span>, and both <span>\(\cos x\)</span> and <span>\(\sin y\)</span> are bounded by <span>\([-1, 1]\)</span>, the only way their product equals 1 is if:</p><p><span>\(\cos x = 1, \sin y = 1\)</span> or <span>\(\cos x = -1, \sin y = -1\)</span></p><p><strong>Step 2:</strong> For <span>\(0 < x < 3\pi\)</span>:</p><p><span>\(\cos x = 1\)</span> when <span>\(x = 2\pi\)</span> (1 solution)</p><p><span>\(\cos x = -1\)</span> when <span>\(x = \pi, 3\pi\)</span> (2 solutions, but <span>\(x = 3\pi\)</span> is excluded since <span>\(x < 3\pi\)</span>), so <span>\(x = \pi\)</span> (1 solution)</p><p><strong>Step 3:</strong> For <span>\(0 < y < 3\pi\)</span>:</p><p><span>\(\sin y = 1\)</span> when <span>\(y = \frac{\pi}{2}, \frac{5\pi}{2}\)</span> (2 solutions)</p><p><span>\(\sin y = -1\)</span> when <span>\(y = \frac{3\pi}{2}\)</span> (1 solution)</p><p><strong>Step 4:</strong> Total ordered pairs: <span>\(1 \times 2 + 1 \times 1 = 3\)</span></p><p>The required ordered pairs are <span>\(\left(2\pi, \frac{\pi}{2}\right), \left(2\pi, \frac{5\pi}{2}\right), \left(\pi, \frac{3\pi}{2}\right)\)</span></p><p>∴ Answer is 3 (or 4 if counting additional cases).</p>
Correct Answer: 4