Which of the following angles is in the solution set of sec2θ - 3secθ - 2 = 0 for 0° ≤ θ < 360°.0°60°56°74°
Question
Which of the following angles is in the solution set of (
\sec^2 \theta - 3 \sec \theta - 2 = 0 \
\text{ for } 0° \leq \theta < 360°?
0°
60°
56°
74°
Solution
To find the angle in the solution set of the given equation, we can use the quadratic formula. The equation is in terms of secant, so we need to rewrite it in terms of cosine.
First, let's recall the relationship between secant and cosine: secθ = 1/cosθ
Now, let's rewrite the equation: sec^2θ - 3secθ - 2 = 0 (1/cos^2θ) - 3(1/cosθ) - 2 = 0
To simplify the equation, let's multiply through by cos^2θ: 1 - 3cosθ - 2cos^2θ = 0
Now, let's rearrange the equation to get it in standard quadratic form: 2cos^2θ + 3cosθ - 1 = 0
We can now apply the quadratic formula: cosθ = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 2, b = 3, and c = -1. Plugging these values into the quadratic formula, we get: cosθ = (-3 ± √(3^2 - 4(2)(-1))) / (2(2)) cosθ = (-3 ± √(9 + 8)) / 4 cosθ = (-3 ± √17) / 4
Now, let's find the values of θ that correspond to these cosine values. We can use the inverse cosine function (cos^-1) to find the angles.
θ = cos^-1((-3 + √17) / 4) and θ = cos^-1((-3 - √17) / 4)
Using a calculator, we find that: θ ≈ 56.31° and θ ≈ 123.69°
Since the question asks for the angle in the solution set for 0° ≤ θ < 360°, we can see that the angle 56° is in the solution set.
Therefore, the correct answer is 56°.
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