The given equation is 2sin^2(θ) + 5sin(θ) = -2.

First, we can rewrite this equation in the standard quadratic form: 2sin^2(θ) + 5sin(θ) + 2 = 0.

This is a quadratic equation in sin(θ), so we can solve it using the quadratic formula: sin(θ) = [-b ± sqrt(b^2 - 4ac)] / (2a).

Here, a = 2, b = 5, and c = 2.

Substituting these values into the quadratic formula gives: sin(θ) = [-5 ± sqrt((5)^2 - 4*2*2)] / (2*2).

Solving this gives: sin(θ) = -1 or sin(θ) = -2.

However, the range of the sine function is -1 ≤ sin(θ) ≤ 1, so sin(θ) = -2 has no solution.

Therefore, the only possible value for θ is when sin(θ) = -1.

(a) For all degree solutions, θ = 180° + 360°k, where k is any integer.

(b) For 0° ≤ θ

Question

The given equation is 2sin^2(θ) + 5sin(θ) = -2.

First, we can rewrite this equation in the standard quadratic form: 2sin^2(θ) + 5sin(θ) + 2 = 0.

This is a quadratic equation in sin(θ), so we can solve it using the quadratic formula: sin(θ) = [-b ± sqrt(b^2 - 4ac)] / (2a).

Here, a = 2, b = 5, and c = 2.

Substituting these values into the quadratic formula gives: sin(θ) = [-5 ± sqrt((5)^2 - 4*2*2)] / (2*2).

Solving this gives: sin(θ) = -1 or sin(θ) = -2.

However, the range of the sine function is -1 ≤ sin(θ) ≤ 1, so sin(θ) = -2 has no solution.

Therefore, the only possible value for θ is when sin(θ) = -1.

(a) For all degree solutions, θ = 180° + 360°k, where k is any integer.

(b) For 0° ≤ θ < 360°, the only solution is θ = 180°.

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