<p>In the interval \(\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\) the equation \(\log_{\sin}(\cos 2\theta) = 2\) has</p>
Step-by-Step Solution
Key Concept: Convert the logarithmic equation log_sin(θ)(cos 2θ) = 2 to exponential form: sin²(θ) = cos 2θ, then use cos 2θ = 1 - 2sin²(θ) to create a quadratic in sin²(θ). The base sin(θ) must satisfy 0 < sin(θ) < 1 and sin(θ) ≠ 1.
<p><strong>Step 1:</strong> For log<sub>sin(θ)</sub>(cos 2θ) = 2 to be defined, we need:</p><ul><li>sin(θ) > 0 and sin(θ) ≠ 1 (base conditions)</li><li>cos 2θ > 0 (argument conditions)</li><li>θ ∈ [-π/2, π/2]</li></ul><p><strong>Step 2:</strong> Convert to exponential form: sin²(θ) = cos 2θ</p><p><strong>Step 3:</strong> Substitute cos 2θ = 1 - 2sin²(θ):</p><p>sin²(θ) = 1 - 2sin²(θ)</p><p>3sin²(θ) = 1</p><p>sin²(θ) = 1/3</p><p>sin(θ) = ±1/√3</p><p><strong>Step 4:</strong> Apply domain restrictions:</p><ul><li>sin(θ) = 1/√3 ✓ (positive, less than 1, and θ ∈ (0, π/2) satisfies sin θ > 0)</li><li>sin(θ) = -1/√3 ✗ (negative, violates sin(θ) > 0)</li></ul><p><strong>Step 5:</strong> Verify cos 2θ > 0: cos 2θ = sin²(θ) = 1/3 > 0 ✓</p><p>∴ The equation has <strong>exactly 1 solution</strong> in the given interval.</p>
Correct Answer: A