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Solve the following equation for angles between 0° and 360°: 1 + sinθ - 2cos²θ = 0

A. 45°
B. 90°
C. 180°
D. 270°

User Lenniekid
by
8.5k points

1 Answer

3 votes

Final answer:

The equation 1 + sinθ - 2cos²θ = 0 is solved by converting the cosine term into a quadratic equation in sine, then factoring to get sinθ = 1/2 and sinθ = -1, resulting in the angles 30°, 150°, and 270°.

Step-by-step explanation:

To solve the equation 1 + sinθ - 2cos²θ = 0 for angles between 0° and 360°, we first need to recognize that cos²θ can be expressed in terms of sine using the Pythagorean identity: cos²θ = 1 - sin²θ. Substituting into the original equation, we get:

1 + sinθ - 2(1 - sin²θ) = 0

Simplifying, we have:

1 + sinθ - 2 + 2sin²θ = 0

So that reduces to:

2sin²θ + sinθ - 1 = 0

This is a quadratic equation in terms of sinθ which can be solved using the quadratic formula, factoring, or completing the square. If we factor the equation, we get:

(2sinθ - 1)(sinθ + 1) = 0

Setting each factor equal to zero gives us the potential solutions:

sinθ = 1/2 or sinθ = -1

Recalling the unit circle values, we find that sinθ = 1/2 corresponds to the angles 30° and 150°, and sinθ = -1 corresponds to the angle 270°. Thus, the correct answers are 30°, 150°, and 270°.

None of the provided options (45°, 90°, 180°, 270°) exactly match all correct solutions, but 270° is one valid solution within the given range. If the question expects a single answer, we need to clarify that multiple solutions exist for the given equation.

User Royston
by
8.4k points
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