solve

Question

3sin^2(x)=5sin(x)+2 (0≤x<π)

Answer 1

3sin^2(x)=5sin(x)+2
for readability, let u = sin(x)

3u^2 - 5u - 2 = 0
(u-2)(3u+1) = 0

so, we need

sin(x) = 2 -- ain't likely, or
sin(x) = -1/3

now, sin(x) >= 0 for 0≤x<π, so there is no solution

Answer 2

3sin^2 x - 5sin x - 2 = 0
(3sinx + 1)(sinx - 2) = 0
sinx = -1/3 or sinx = 2, the last part is not possible

if sinx = -1/3
x must be in quadrants III or IV , but 0≤x ≤ π
so there is no solution in your given domain