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 (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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We write 2cos x + 1 = 0 as:
cos x + cos x + 1 = 0
We'll substitute the values 1 = cos 0 and 0 = cos pi/2.
We'll subtract cos 0 both sides:
cos x + cos x = cos pi/2 - cos 2pi (1)
We'll transform the sum and the difference of cosines in products:
cos x + cos x = 2 cos [(x+x)/2]* cos [(x-x)/2]
cos x + cos x = 2cos x*cos 0 (2)
cos pi/2 - cos 2pi = 2 sin pi/4*sin (2pi- pi/2)/2
cos pi/2 - cos 2pi = 2sin pi/4*sin 3 pi/4 (3)
We'll substitute (2) and (3) in (1):
2cos x*cos 0 = 2sin pi/4*sin 3 pi/4
We'll divide by 2:
cos x*cos 0 = sin pi/4*sin 3 pi/4
We'll put cos 0 = 1
cos x*1= sin pi/4*sin 3 pi/4
sin pi/4 = cos (pi/2 - pi/4)
sin pi/4 = cos pi/4
cos x = cos pi/4
x = arccos (cos pi/4) + 2k*pi
x = pi/4
sin 3pi/4 = cos (pi/2 - 3pi/4)
x = cos (-pi/4)
Since the cosine function is even, we'll have:
cos (-pi/4) =cos (pi/4)
x = pi/4
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