How do you solve Cos(pi/2 * x - 1) = 0cos(π2⋅x−1)=0?

1 Answer
Cesareo R.
May 13, 2016

x =(2 + (1 + 4 k) pi)/pi, k = {0,pm 1,pm 2,..}x=2+(1+4k)ππ,k={0,±1,±2,..}

Explanation:

The equation cos(theta)=0cos(θ)=0 verifies for all theta = pi/2+2k piθ=π2+2kπ with k = {0,pm 1,pm 2,..}k={0,±1,±2,..}
Then pi/2x-1π2x−1 must be of the form pi/2+2k piπ2+2kπ. Equating
pi/2x-1=pi/2+2k piπ2x−1=π2+2kπ Solving for xx we get
x =(2 + (1 + 4 k) pi)/pi, k = {0,pm 1,pm 2,..}x=2+(1+4k)ππ,k={0,±1,±2,..}

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