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`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` .
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