Answer:
We have,
sin θ = 1
⇒ sin θ = sin \(\frac{π}{2}\)
θ = mπ + (-1)\(^{m}\) ∙ \(\frac{π}{2}\), m ∈ Z, [Since, the general solution of sin θ = sin ∝ is given by θ = nπ + (-1)\(^{n}\) ∝, n ∈ Z.]
Now, if m is an even integer i.e., m = 2n (where n ∈ Z) then,
θ = 2nπ + \(\frac{π}{2}\)
⇒ θ = (4n + 1)\(\frac{π}{2}\)
BTW u wrote your question in the wrong 1 (u need to put this in mathematics!)
