Assume a≠kπ as the general solution of sin(a)=0. If k is any integer and zero is an integer then k=0 is a possible value for k. But if k=0 then, since a≠kπ on our assumption, a≠0 is a solution for sin(a)=0. In a right-angled triangle sine=opposite side/hypotenuse by definition. So if sin(a)=0 the opposite side length must be permitted to be zero, since no value of the hypotenuse would make the quotient zero. As the length of the opposite side approaches zero, the angle, a, gets smaller. In the limit, the opposite side length becomes zero, and the angle a must also be zero, so a=0 is a solution to sin(a)=0, so contradicting the implication that a≠0. This would eliminate k=0 as an integer.
Also, sin(π-a)=sin(a). When sin(a)=0, sin(π)=0. When k=1, if we assume a≠π is a solution we have the contradiction that sin(π)=0. And because sine is a periodic function sin(2π+a)=sin(a) so the argument can be extended to sin(2π), sin(2π+π)=sin(3π)= etc. Therefore we are led to the conclusion that a=kπ is a solution when sin(a)=0.