If 2 sin² θ + 3 sin θ - 2 = 0, where 0

Question

If 2 sin² θ + 3 sin θ - 2 = 0, where 0 < θ < 90°, then find the value of θ.

Accepted Answer

Given the equation: 2 sin² θ + 3 sin θ - 2 = 0

Step 1: Let x = sin θ.
The equation becomes: 2x² + 3x - 2 = 0

Step 2: Solve the quadratic equation:
Using the quadratic formula x = [-b ± sqrt(b² - 4ac)] / 2a
Here, a = 2, b = 3, c = -2
Discriminant = 3² - 4*2*(-2) = 9 + 16 = 25

x = [-3 ± 5] / (2*2) = [-3 ± 5] / 4

Two roots:
x₁ = (-3 + 5)/4 = 2/4 = 0.5
x₂ = (-3 - 5)/4 = -8/4 = -2 (not possible since sin θ must be between -1 and 1)

Step 3: So, sin θ = 0.5

Step 4: Find θ in (0°, 90°) such that sin θ = 0.5
θ = 30°

Therefore, the value of θ is 30°.

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