Question

The general solution of sinx=1​

Answer 1

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!)