If 2 sin² θ + 3 sin θ − 2 = 0, (0 < θ < 90°), then the value of θ is:

Correct — A. The stem, 2 sin²θ + 3 sin θ − 2 = 0, is a quadratic in sin θ, not in θ. Substitute and factorise.

Let s = sin θ: 2s² + 3s − 2 = 0
Split the middle term: 2s² + 4s − s − 2 = 0
2s(s + 2) − 1(s + 2) = 0
(2s − 1)(s + 2) = 0

So s = 1⁄2 or s = −2. A sine can never be −2, so that root goes.

sin θ = 1⁄2 with 0 < θ < 90° gives θ = 30° → option (a), the image showing 30°.

Why the others are wrong

• (b)
  15° is the half-of-30° trap. sin 15° ≈ 0.259, and putting it back gives 2(0.067) + 3(0.259) − 2 ≈ −1.09, not 0.
• (c)
  45° has sin θ = 1⁄√2 ≈ 0.707. The left side becomes 2(0.5) + 3(0.707) − 2 ≈ +1.12, so the equation is not satisfied.
• (d)
  60° is the angle for sin θ = √3⁄2, not 1⁄2. Substituting gives 2(0.75) + 3(0.866) − 2 ≈ +2.10, well clear of 0.

Any equation built only from one trigonometric ratio can be solved as ordinary algebra: name the ratio s, solve for s, then convert s back to an angle.

The extra step trigonometry adds is a range filter. For a real angle −1 ≤ sin θ ≤ 1, so a root outside that band is discarded no matter how sound the algebra was.

Only then does the standard-angle table come in, and only the acute quadrant is in play because the stem restricts θ to 0 < θ < 90°.

Key facts

• sin 30° = 1⁄2, sin 45° = 1⁄√2 and sin 60° = √3⁄2.
• For any real angle, −1 ≤ sin θ ≤ 1, so a root of −2 is rejected.
• 2s² + 3s − 2 factorises as (2s − 1)(s + 2).

Common traps

• Stopping at sin θ = 1⁄2 and marking a value rather than an angle.
• Keeping the root s = −2 because the factorisation produced it.
• Treating the equation as quadratic in θ and reaching for the discriminant in degrees.

No directly related past PYQ was found.