If 2 cos2θ + 11 sinθ - 7 = 0, then the value of sinθ, where [0 ≤ θ ≤ \(\frac{{\rm{\pi }}}{2}\)
2cos2θ + 11 sinθ - 7 = 0
2(1 - sin2θ) + 11 sinθ - 7 = 0
2sin2θ - 11 sinθ + 5 = 0
2sin2 θ - 10 sin θ - sin θ + 5 = 0
2sinθ (sinθ - 5) - 1 (sinθ - 5) = 0
(2sinθ - 1) (sinθ - 5) = 0
∴ sinθ $( = \frac{1}{2})$
[As sinθ = 5 is not possible]
[As we know sinθ = 0 ≤ θ ≤ $(\frac{{\rm{\pi }}}{2})$]