2cosx + 1 = 0  Find x values for the interval [0, 2pi].

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You need to solve for `x in [0,2pi]` the equation `2cos x + 1 = 0` , such that:

`2cos x = -1 => cos x = -1/2 => {( cos x = cos(pi - pi/3)),( cos x = cos (pi + pi/3)):}`

`{(cos x = cos(2pi/3)),(cos x = cos...

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You need to solve for `x in [0,2pi]` the equation `2cos x + 1 = 0` , such that:

`2cos x = -1 => cos x = -1/2 => {( cos x = cos(pi - pi/3)),( cos x = cos (pi + pi/3)):}`

`{(cos x = cos(2pi/3)),(cos x = cos (4pi/3)):} => {(x = (2pi)/3),(x = (4pi)/3):}`

Hence, evaluating the solutions to the given equation, under the given conditions, yields `x = (2pi)/3` , `x = (4pi)/3.`

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