# Use half angle formulas. Find all solutions in the interval [0, 2pi). cos x/2 + cos x = 0

*print*Print*list*Cite

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

To solve, apply the half-angle identity ` cos(theta/2)=+-sqrt((1+costheta)/2)` .

`+-sqrt((1+cosx)/2)+cosx=0`

Then, isolate the radical in our equation.

`+-sqrt((1+cosx)/2) = -cosx`

Take the square root of both sides to eliminate the square root.

`(1+cosx)/2 = cos^2x`

`1/2+1/2cosx = cos^2x`

Since the result is a quadratic equation, to solve it, one side should be zero. So, move the terms at the left to the right side.

`0=cos^2x-1/2cosx-1/2`

Next, factor the right side.

0=(cosx - 1)(cosx + 1/2)

Then, set each factor equal to zero. And solve for x. Here, we have to take note that the values of x should be within the interval [0,2pi].

For the first factor:

`cosx - 1 = 0`

`cosx = 1`

`x = 0`

For the second factor:

`cosx +1/2 = 0`

`cosx = -1/2`

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

Since the original equation contains half-angle, it is necessary to plug-in the values of x that we got to determine which of them satisfy the original equation.

When x = 0, we would have:

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

`cos(0)+cos(0)=0`

`1+1=0`

`2!=0`

When x = (2pi)/3, we would have:

`cos(((2pi)/3)/2)+cos((2pi)/3)=0`

`cos(pi/3)+cos((2pi)/3)=0`

`1/2+(-1/2)=0`

`0=0`

When x = (4pi)/3, we would have:

`cos(((4pi)/3)/2)+cos((4pi)/3)=0`

`cos((2pi)/3)+cos((4pi)/3)=0`

`-1/2+(-1/2)=0`

`-1!=0`

Notice that among the three values of x, only (2pi)/3 satisfy the original equation.

**Therefore, the solution of `cos(x/2)+cosx=0` in the interval `[0,2pi)` is `x=(2pi)/3` .**