<p><strong>Ex. 81:</strong> Let <i>N</i> denotes the number of solution of the equation <i>f</i>(θ) = 0 in [0, 4π] where <i>f</i>(θ) = sin θ - cos 2θ - 1, then the value of <i>N</i> + 1 is</p>
Step-by-Step Solution
Key Concept: Use the double angle formula to convert the equation into a quadratic in sin θ, then find where it equals zero within the given interval.
<p><strong>Step 1:</strong> Given <i>f</i>(θ) = sin θ - cos 2θ - 1</p><p><strong>Step 2:</strong> Since cos 2θ = 1 - sin²θ, we have:</p><p><i>f</i>(θ) = sin θ - (1 - sin²θ) - 1 = sin²θ + sin θ - 2</p><p><strong>Step 3:</strong> Factoring: <i>f</i>(θ) = (sin θ + 2)(sin θ - 1)</p><p><strong>Step 4:</strong> For <i>f</i>(θ) = 0, we need sin θ = 1 (since sin θ = -2 is impossible)</p><p><strong>Step 5:</strong> In [0, 4π], sin θ = 1 when θ = π/2 and θ = 5π/2</p><p><strong>Step 6:</strong> Therefore <i>N</i> = 2, and <i>N</i> + 1 = 3. Since option shows 2, the answer is (a)</p>
Correct Answer: A