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 step by step, we will analyze each inequality separately and then find the intersection of their solutions.

### Step 1: Solve the first inequality
The first inequality we need to solve is:
\[ 2 \sin^2 x + 3 \sin x - 2 > 0 \]

#### Step 1.1: Factor the quadratic
Let's set \( y = \sin x \). The inequality becomes:
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