When 0° < θ < 90°, solve the following equation:

2 cos² θ + sin θ - 2 = 0

Given,

2 cos² θ + sin θ - 2 = 0

On Solving,

⇒ 2(1 - sin² θ) + sin θ - 2 = 0

= 2 - 2 sin² θ + sin θ - 2 = 0

= sin θ -2 sin² θ = 0

= sin θ (1 - 2 sin θ) = 0

So, either sin θ = 0 or 1 - 2 sin θ = 0

If, sin θ = 0
sin θ = sin 0°
θ = 0°.

Given, θ > 0° hence, θ = 0° is not possible.

∴ 1 - 2 sin θ = 0

⇒ 1 = 2 sin θ

⇒ sin θ = $\dfrac{1}{2}$

⇒ sin θ = 30°.

Hence, the value of θ = 30°.