Question

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

Answer 1

`(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`

Answer 2

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`