Let `2sin^2x+3sinx-2>0a n dx^2-x-2<0(x` is measured in radians). Then `x` lies in the interval `(pi/6,(5pi)/6)` (b) `(-1,(5pi)/6)` `(-1,2)` (d) `(pi/6,2)`

Explanation

To solve the given inequalities \(2\sin^2 x + 3\sin x - 2 > 0\) and \(x^2 - x - 2 < 0\), we will go through the following steps:

### Step 1: Solve the first inequality \(2\sin^2 x + 3\sin x - 2 > 0\)

1. **Factor the quadratic expression**:
We can rewrite the quadratic as:
\[
2\sin^2 x + 4\sin x - \sin x - 2 > 0
...