Question

Solve for θ in cos(θ) = √3/2 on the interval 0 < θ < 2π.

a) θ = π/3
b) θ = 2π/3
c) θ = 4π/3
d) θ = 5π/6

Answer 1

The angles where cos(θ) = √3/2 in the interval 0 < θ < 2π are π/6 and 11π/6, corresponding to the first and fourth quadrants of the unit circle where the cosine is positive.

Step-by-step explanation:

To solve for θ in cos(θ) = √3/2 within the interval 0 < θ < 2π, we can refer to the unit circle where the cosine of an angle represents the x-coordinate of the point on the unit circle corresponding to that angle.

Since cos(θ) = √3/2, we look for the angles where the x-coordinate is √3/2. Remembering the special triangles and their angles (specifically the 30-60-90 triangle), we recognize that the cosine of π/6 (or 30°) is √3/2. However, because cosine is positive in both the first and fourth quadrants, we must find the corresponding angle in the fourth quadrant as well.

The second angle is 2π - π/6, which simplifies to 11π/6. Thus, the angles are:

1. θ = π/6
2. θ = 11π/6