Trigonometry & Inverse Trigonometry
Trigonometric Inequalities
Grade 11
Question:
<p>If \(0 \leq \theta \leq 2\pi\) and \(2\sin^2 \theta - 5\sin \theta + 2 > 0\), then find the range of \(\theta\).</p>
<p>(a) \(\left(0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, 2\pi\right)\)</p>
<p>(b) \(\left(0, \frac{5\pi}{6}\right) \cup (\pi, 2\pi)\)</p>
<p>(c) \(\left(0, \frac{\pi}{6}\right) \cup \left(\pi, 2\pi\right)\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Convert the trigonometric inequality into a polynomial inequality, then solve for the angle using the unit circle.
<p><strong>Solution:</strong> Let \(y = \sin \theta\). Then \(2y^2 - 5y + 2 > 0\) factors as \((2y - 1)(y - 2) > 0\). This gives \(y < \frac{1}{2}\) or \(y > 2\). Since \(\sin \theta \leq 1\), we have \(y > 2\) is impossible. Thus \(\sin \theta < \frac{1}{2}\). For \(\theta \in [0, 2\pi]\), this means \(\theta \in \left(0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, 2\pi\right)\).</p>
Correct Answer: a