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

Expert Answers
lemjay eNotes educator| Certified Educator

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

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


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.


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:





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





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





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` .