Trigonometry & Inverse Trigonometry
Trigonometric Inequalities
Grade 11
Question:
<p>Let \(2\sin^2 x + 3\sin x - 2 \geq 0\) and \(x^2 - x - 2 < 0\) where \(x\) is measured in radians. Find the interval in which \(x\) lies.</p>
<p>(a) \(\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)\)</p>
<p>(b) \(\left(-1, \frac{5\pi}{6}\right)\)</p>
<p>(c) \((-1, 2)\)</p>
<p>(d) \(\left(\frac{\pi}{6}, 2\right)\)</p>
Step-by-Step Solution
Key Concept: Solve the trigonometric inequality and the polynomial inequality separately, then find their intersection.
<p><strong>Solution:</strong> For \(2\sin^2 x + 3\sin x - 2 \geq 0\), let \(y = \sin x\). Then \(2y^2 + 3y - 2 \geq 0\) factors as \((2y - 1)(y + 2) \geq 0\), giving \(y \leq -2\) or \(y \geq \frac{1}{2}\). Since \(-1 \leq \sin x \leq 1\), we have \(\sin x \geq \frac{1}{2}\). For \(x^2 - x - 2 < 0\), we factor as \((x - 2)(x + 1) < 0\), giving \(-1 < x < 2\). Combining: \(\sin x \geq \frac{1}{2}\) and \(-1 < x < 2\). In the interval \((-1, 2)\), \(\sin x \geq \frac{1}{2}\) when \(x \in \left(\frac{\pi}{6}, \frac{5\pi}{6}\right)\) (within the radians context).</p>
Correct Answer: a