<p>The number of solutions of the equation \(8\tan^2\theta + 9 = 6\sec\theta\) in the interval \(\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\) is</p>
Step-by-Step Solution
Key Concept: Convert the equation to a quadratic in sec(θ) using the identity tan²(θ) = sec²(θ) - 1, then solve and check which solutions are valid within the given domain.
<p><strong>Step 1:</strong> Use the identity tan²(θ) = sec²(θ) - 1 to rewrite the equation.</p><p>8(sec²(θ) - 1) + 9 = 6sec(θ)</p><p>8sec²(θ) - 8 + 9 = 6sec(θ)</p><p>8sec²(θ) - 6sec(θ) + 1 = 0</p><p><strong>Step 2:</strong> Let x = sec(θ). Solve the quadratic: 8x² - 6x + 1 = 0</p><p>Using the quadratic formula: x = (6 ± √(36 - 32))/16 = (6 ± 2)/16</p><p>x = 1/2 or x = 1</p><p><strong>Step 3:</strong> Check validity. Since sec(θ) must satisfy |sec(θ)| ≥ 1, the value sec(θ) = 1/2 is invalid.</p><p><strong>Step 4:</strong> For sec(θ) = 1, we have cos(θ) = 1, giving θ = 0.</p><p>In the interval (-π/2, π/2), θ = 0 is the only solution.</p><p>∴ Answer: <strong>1</strong> (Option A)</p>
Correct Answer: A