First we will rearrange the expression,

`cos(x/2)+cos(x) = 0`

By using trignometric identities,

`cos(x)+cos(x/2) = 2cos((x+x/2)/2)cos((x-x/2)/2) = 0`

This gives,

`cos((3x)/4)cos(x/4) = 0`

this gives two primary solutions,` `` ``cos((3x)/4) = 0` or `cos(x/4) = 0`

(i) `cos((3x)/4) = 0`

this gives, `(3x)/4 = pi/2`

and `x...

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First we will rearrange the expression,

`cos(x/2)+cos(x) = 0`

By using trignometric identities,

`cos(x)+cos(x/2) = 2cos((x+x/2)/2)cos((x-x/2)/2) = 0`

This gives,

`cos((3x)/4)cos(x/4) = 0`

this gives two primary solutions,` ` ` ``cos((3x)/4) = 0` or `cos(x/4) = 0`

(i) `cos((3x)/4) = 0`

this gives, `(3x)/4 = pi/2`

and `x = (2pi)/3`

This is the primary solution, the general solution for cosine is,

`x = +-(2pi)/3 +2npi` where n is any integer,

n = 0, you get `x = (-2pi)/3 and (2pi)/3`

n =1, you get `x = (4pi)/3 and (8pi)/3`

from these answers only values withing the range are `(2pi)/3 and (4pi)/3`

(ii) `cos(x/4) = 0`

this gives `x/4 = pi/2`

and `x = 2pi `

This is the primary solution, the general solution for cosine is,

`x = +-2pi +2npi`

n =0. you get `x = -2pi and +2pi`

n = 1, you get, `x = 0 and +4pi`

Therefore the answers within our range are `x = 0 and 2pi`

Therefore th complete set of answers are `(2pi)/3, (4pi)/3,0 and 2pi`