The set of all `x` in the interval `[0,pi]` for which `2sin^2x-3sinx+1geq0` is______

Explanation

To solve the inequality \(2\sin^2 x - 3\sin x + 1 \geq 0\) for \(x\) in the interval \([0, \pi]\), we can follow these steps:

### Step 1: Substitute \( \sin x \) with \( t \)
Let \( t = \sin x \). The inequality then becomes:
\[ 2t^2 - 3t + 1 \geq 0 \]

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