MHT CET · Maths · Trigonometric Equations
Let \(2 \sin ^2 x+3 \sin x-2\gt0\) and \(x^2-x-2 \lt 0\) ( \(x\) is measured in radians).
Then \(x\) lies in the interval
- A \(\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)\)
- B \(\left(-1, \frac{5 \pi}{6}\right)\)
- C \((-1,2)\)
- D \(\left(\frac{\pi}{6}, 2\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{\pi}{6}, 2\right)\)
Step-by-step Solution
Detailed explanation
\(\therefore 2 \sin ^2 x+3 \sin x-2\gt0 \)
\( (2 \sin x-1)(\sin x+2)\gt0 \)
\( \Rightarrow 2 \sin x-1\gt0 \ldots[\because \sin x+2\gt\) \(0 \text { and } x \in \mathrm{R}] \)
\( \Rightarrow \sin x\gt\frac{1}{2} \)
\( \Rightarrow x \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)\)
Also\(, x^2-x-2 \lt 0 \)
\( \Rightarrow(x-2)(x+1) \lt 0 \)
\( \Rightarrow-1 \lt x \lt 2\)
Since \(2 \lt \frac{5 \pi}{6}\)
\(\therefore x\) must lie in \(\left(\frac{\pi}{6}, 2\right)\)
\( (2 \sin x-1)(\sin x+2)\gt0 \)
\( \Rightarrow 2 \sin x-1\gt0 \ldots[\because \sin x+2\gt\) \(0 \text { and } x \in \mathrm{R}] \)
\( \Rightarrow \sin x\gt\frac{1}{2} \)
\( \Rightarrow x \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)\)
Also\(, x^2-x-2 \lt 0 \)
\( \Rightarrow(x-2)(x+1) \lt 0 \)
\( \Rightarrow-1 \lt x \lt 2\)
Since \(2 \lt \frac{5 \pi}{6}\)
\(\therefore x\) must lie in \(\left(\frac{\pi}{6}, 2\right)\)
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