The only possible value of sinθ satisfying 2si$n^2$θ − 5sinθ + 2 = 0 is:

Factorising: (2sinθ − 1)(sinθ − 2) = 0, so sinθ = 1/2 or sinθ = 2. A sine never leaves the interval −1 to 1, so the root 2 must be discarded and only sinθ = 1/2 survives. (5/2 is the sum of the two roots, not a root.)