Solve the equation on the interval
0
≤
𝜃
≤
2
𝜋

1
−
cos
⁡
𝜃
=
1
2

Question

Solve the equation on the interval 
0
≤
𝜃
≤
2
𝜋

1
−
cos
⁡
𝜃
=
1
2

BlackTom AI · Accepted Answer

We start by rewriting the equation in a form we can analyze clearly.

Given: 1 − cos(θ) = 1/2 on the interval 0 ≤ θ ≤ 2π.

First, isolate cos(θ): subtract 1 from both sides to get −cos(θ) = 1/2 − 1 = −1/2, then multiply by −1 to obtain cos(θ) = 1/2.

Next, recall the basic angles where cosine equals 1/2. Cosine is 1/2 at θ values of θ = π/3 and θ = −π/3 within the standard circle.

Since cosine is 2π-periodic, all solutions are given by θ = ±π/3 + 2kπ for integers k.

To find the solutions within the interval [0, 2π], we generate the specific angles:
- θ = π/3 (which lies in [0, 2π])
- θ = −π/3 is equivalent to θ = 2π − π/3 = 5π/3, which also lies in [0, 2π].

We should also check the endpoints 0 and 2π just to confirm they are not solutions: cos(0) = 1, so 1 − cos(0) = 0 ≠ 1/2; cos(2π) = 1, so 1 − cos(2π) = 0 ≠ 1/2. Therefore, neither endpoint is a solution.

Thus, the solutions on 0 ≤ θ ≤ 2π are the angles where cos(θ) = 1/2, specifically θ = π/3 and θ = 5π/3.

If the answer field lists these two angles as the solution, that aligns with the correct set of solutions for the given equation on the specified interval.

(H) Precalculus - Pak: 24-25_YR Ch 7 Quiz (7.1-7.4)

Solve the equation on the interval 0 ≤ 𝜃 ≤ 2 𝜋 1 − cos ⁡ 𝜃 = 1 2

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