2cosx + 1 = 0 find x values for the interval [0, 2pi]
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bookB.S. from Mahatma Gandhi University Kottayam, Kerala, India
bookM.S. from University of Kerala , India
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We have to find the value of x for `2cosx+1=0` in the interval `[0,2\pi]`
So,
`2cosx+1=0`
In other words, `2cosx=-1`
`cosx=\frac{-1}{2}`
The solution will be of the form:
`x=\pm cos^{-1}(\frac{-1}{2})+2\pi n` where n=0,1,2,. . .
For example, `x=\pm \frac{2\pi}{3} +2\pi n , \ n=0,1,2,...`
When ,
`n=0 : x=\pm \frac{2\pi}{3}`
`n=1:\ x= \frac{2\pi}{3}+2\pi = \frac{8\pi}{3}`
or , `x=\frac{-2\pi}{3}+2\pi=\frac{4\pi}{3}`
However, we have to find the angle in the interval `[0,2\pi]`
As a result, the answer is:
`x=\frac{2\pi}{3} \ and \ x=\frac{4\pi}{3}`
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calendarEducator since 2010
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The equation 2cosx + 1 = 0 has to be solved for values of x in the set `[0,2pi]`
`2*cos x + 1 = 0`
=> `2*cos x = -1`
=> `cos x = -1/2`
=> `x = 2*n*pi +-cos^-1(-1/2)`
(The entire section contains 4 answers and 390 words.)
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calendarEducator since 2013
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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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