Solve cos2 θ – sin θ – \(\frac{1}{4}\) = 0 (0° < θ < 360°)

1.	Solve cos2 θ – sin θ – \(\frac{1}{4}\) = 0 (0° < θ < 360°)
Answer» cos² θ – sin θ - \(\frac{1}{4}\) = 0 ⇒ 1 – sin² θ – sin θ – \(\frac{1}{4}\) = 0 ⇒ 4 sin² θ + 4 sin θ – 3 = 0 ⇒ (2 sin θ + 3) (2 sin θ – 1) = 0 ⇒ 2 sin θ + 3 = 0 or 2 sin θ – 1 = 0 ⇒ sin θ = - \(\frac{3}{2}\) or sin θ = \(\frac{1}{2}\) ⇒ θ = 30°, 150°. Since \|sin θ\| = \(\frac{3}{2}\) is > 1, the value sin θ = - \(\frac{3}{2}\) is inadmissible. ∴ θ = 30°, 150°.

Answer»

cos² θ – sin θ - \(\frac{1}{4}\) = 0

⇒ 1 – sin² θ – sin θ – \(\frac{1}{4}\) = 0

⇒ 4 sin² θ + 4 sin θ – 3 = 0

⇒ (2 sin θ + 3) (2 sin θ – 1) = 0

⇒ 2 sin θ + 3 = 0 or 2 sin θ – 1 = 0

⇒ sin θ = - \(\frac{3}{2}\) or sin θ = \(\frac{1}{2}\)

⇒ θ = 30°, 150°.

Since |sin θ| = \(\frac{3}{2}\) is > 1, the value sin θ = - \(\frac{3}{2}\) is inadmissible.

∴ θ = 30°, 150°.

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