Given that $ cos(θ) = -frac{1}{2} $, find the general solutions for $ θ $ in the unit circle.
Answer 1
Emma Johnson
To solve for $ θ $ such that $ \cos(θ) = -\frac{1}{2} $, we need to find all angles in the unit circle where the cosine value is $ -\frac{1}{2} $. The cosine function is negative in the second and third quadrants.
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The reference angle for $ \cos(θ) = \frac{1}{2} $ is $ \frac{\pi}{3} $. Therefore, the general solutions in the second and third quadrants are:
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$ θ = \pi – \frac{\pi}{3} + 2k\pi = \frac{2\pi}{3} + 2k\pi $
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and
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$ θ = \pi + \frac{\pi}{3} + 2k\pi = \frac{4\pi}{3} + 2k\pi $
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where $ k $ is any integer.
Answer 2
William King
To find $ θ $ such that $ cos(θ) = -frac{1}{2} $, consider the reference angle $ frac{pi}{3} $. The cosine function is negative in the second and third quadrants:
$ θ = frac{2pi}{3} + 2kpi $
and
$ θ = frac{4pi}{3} + 2kpi $
where $ k $ is any integer.
Answer 3
Henry Green
Given $ cos(θ) = -frac{1}{2} $, the solutions are:
$ θ = frac{2pi}{3} + 2kpi $
and
$ θ = frac{4pi}{3} + 2kpi $
where $ k $ is any integer.
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