Solve 2 cos² θ + sin θ - 2 = 0, when 0°

Question

Solve 2 cos² θ + sin θ - 2 = 0, when 0° < θ < 90°

KnowledgeBoat · Accepted Answer

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 = ⇒ sin θ = 30°. Hence, the value of θ = 30°.

Mathematics

When 0° < θ < 90°, solve the following equation:
2 cos² θ + sin θ - 2 = 0

Trigonometric Identities

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Mathematics

When 0° < θ < 90°, solve the following equation:2 cos2 θ + sin θ - 2 = 0

Trigonometric Identities

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When 0° < θ < 90°, solve the following equation:3 cos θ = 2 sin2 θ

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When 0° < θ < 90°, solve the following equation:
2 cos² θ + sin θ - 2 = 0

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
2(sin⁶ θ + cos⁶ θ) - 3(sin⁴ θ + cos⁴ θ) + 1 = 0.

If cot θ + cos θ = m and cot θ - cos θ = n, then prove that (m² - n²)² = 16 mn.

When 0° < θ < 90°, solve the following equation:
3 cos θ = 2 sin² θ

When 0° < θ < 90°, solve the following equation:
sec² θ - 2 tan θ = 0