How do you solve 2cosx+1=0 algebraically?

2 Answers
Jun 14, 2018

(2pi)/3, (4pi)/3 on 0<=x<=2pi

Explanation:

2cosx+1=0

2cosx=-1

cosx=-1/2

cos^-1(-1/2) = x

Use the unit circle to solve:

cos^-1(-1/2) = (2pi)/3, (4pi)/3

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Jun 14, 2018

Given:

2cos(x)+1=0

Subtract 1 from both sides of the equation:

2cos(x)+1color(red)(-1)=0color(red)(-1)

2cos(x)=-1

Divide both sides of the equation by 2:

2/color(red)(2)cos(x)=-1/color(red)(2)

cos(x)=-1/2

Use the inverse cosine function on both sides:

color(red)(cos^-1)(cos(x))=color(red)(cos^-1)(-1/2)

x=cos^-1(-1/2)

Within the domain 0<=x<2pi the above equation is true for 2 values of x:

x = 2/3pi and x = 4/3pi

To represent all of the possible solutions, we add integer multiples of 2pi to both answers:

x = 2/3pi+ 2npi and x = 4/3pi + 2npi, n in ZZ