Question

How do you solve the following equation `cos(x+pi/3)+cosx=0` in the interval [0, 2pi]?

Answer 1

`x=npi+pi/3`

Explanation:

`cos(x+pi/3)+cosx=0` can be expanded to

`cosxcos(pi/3)-sinxsin(pi/3)+cosx=0`

or `cosx/2-sqrt3/2sinx+cosx=0` or

`3/2cosx=sqrt3/2sinx`

`tanx=(3/2)/(sqrt3/2)=3/2xx2/sqrt3=sqrt3=tan(pi/3)`

Hence `x=npi+pi/3`