Let $2 \sin ^2 x+3 \sin x-2\gt0$ and $x^2-x-2 \lt 0$. ( $x$ is measured in radians). The $x$ lies in the…
Let $2 \sin ^2 x+3 \sin x-2\gt0$ and $x^2-x-2 \lt 0$. ( $x$ is measured in radians). The $x$ lies in the interval
- $\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)$
- $\left(-1, \frac{5 \pi}{6}\right)$
- $(-1,2)$
- $\left(\frac{\pi}{6}, 2\right)$
Solution
$2 \sin ^2 x+3 \sin x-2\gt0...(i)$
Let $y=\sin x$
$\therefore \quad$ Equation (i) becomes
$\begin{array}{ll}
& 2 y^2+3 y-2\gt0 \\
\therefore \quad & 2 y^2+4 y-y-2\gt0 \\
\therefore \quad & 2 y(y+2)-1(y+2)\gt0 \\
\therefore \quad & (2 y-1)(y+2)\gt0 \\
\therefore \quad & (2 y-1)\gt0 \text { and }(y+2)\gt0 \quad \text { OR } \\
& (2 y-1) \lt 0 \text { and }(y+2) \lt 0
\end{array}$
$\begin{array}{ll}\therefore \quad & y\gt\frac{1}{2} \text { and } y\gt-2 \\ & y \lt \frac{1}{2} \text { and } y \lt -2 \\ \therefore \quad & \sin x\gt\frac{1}{2} \text { and } \sin x\gt-2 \\ & \sin x \lt \frac{1}{2} \text { and } \sin x \lt -2\end{array}$
$\therefore \quad x\gt\frac{\pi}{6}$ and $\sin x\gt-2$
$\ldots[\because-1 \leq \sin x \leq 1$, second condition is not possible]
$\therefore \quad x\gt\frac{\pi}{6}$ and $\sin x\gt-1 \quad \ldots[\because-1 \leq \sin x \leq 1]$
$\begin{aligned}
& \therefore \quad x \in\left(\frac{\pi}{6}, \infty\right) \\
& \text { Given that } x^2-x-2 \lt 0 \\
& \therefore \quad(x-2)(x+1) \lt 0 \\
& \therefore \quad(x-2) \lt 0 \text { and }(x+1)\gt0 \quad \text { OR } \\
& (x-2)\gt0 \text { and }(x+1) \lt 0 \\
& \therefore \quad x \lt 2 \text { and } x\gt-1 \\
& x\gt2 \text { and } x \lt -1 \\
& \therefore \quad x \in(-1,2)
\end{aligned}$
OR
...(ii) [ $\because$ second condition is not possible]
$\therefore \quad$ from (i) and (ii), we get
$x \in\left(\frac{\pi}{6}, 2\right)$
Asked in: MHT CET 2024 (03 May Shift 1)
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