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The answer above is incomplete. The domain of `cos^(-1)x` is `[0,pi)` . `cos^(-1)0=pi/2` but `cosx=0` if `x=(3pi)/2` also.
The complete solution is `x=pi/2 +npi,x=pi+2npi` for `n in ZZ`
The solution to the equation `cos^2x + cos x = 0` has to be determined.
`cos^2x + cos x = 0`
=> `cos x(cos x + 1) = 0`
=> cos x = 0 and cos x = -1
=> x = `cos^-1 0` and x =`cos^-1(-1)`
=> x = `pi/2 + 2*n*pi` and x = `pi + 2*n*pi`
The solution of the equation cos^2x + cos x = 0 is `x = pi/2 + 2*n*pi` and `x = pi + 2*n*pi`
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