#x# could be either in radians or in degrees.
Let take #x# to be in radians.
One solution of this equation would be #x=sin^-1(a)#
Because #sin^-1# is the inverse trig function for #sin#
This one solution(#x=sin^-1(a)#) is the only one!
Because we know that adding #2pi# to an angle gives an angle equivalent to the former.
This means that we could keep adding #2pi# to this angle and get infinitely more solutions
Letting #A=sin^-1(a)# ,
A more complete solution would be #x=A+(2pi)n#
#ninNN#
#color(green )"Note: "#Multiplying by #n# means that adding any multiple of #2pi# is a solution.
This is however still incomplete.
All complemenatary angles are also a solution.
#color(green)"Recall: "# Complementary angles sums up to #180# or #2pi# Example : #30º+150º=180#
So, #30º# and #150º# are each other's complemenatry angles.
Thus, for every solution #A# , another solution is : #pi-A# , because #A+(pi-A)=pi#
As before, to this other solution we can add multiples of #2pi#
to have equivalent solutions.
Hence, another set of solutions is : #x=pi-A+(2pi)n#
So, a complete solution for #sinx=a# is :
#color(blue)(x=A+(2pi)n)#
and #color(blue)(x=pi-A+(2pi)n)#
To combine the two sets of solutions,
Rearrange #x=pi-A+(2pi)n# to #color(green)(x=-A+(2n+1)pi)#
Since #ninNN# we notice that , #(2n+1)# is always an odd number
Rearrange #x=A+(2pi)n# to #color(green)(x=A+(2n)pi)#
See that, #(2n)# is always an even number
So, the trick is that when we write the two solutions as , #x=(-1)^nA+npi#
When #n# is even, we are using the first solution : #color(orange)(x=A+(2n)pi)#
and when #n# is odd, we are using the second solution : #color(orange)(x=-A+(2n+1)pi)#
NB: when #n# is even #(-1)^n=color(red)1#
and when #n# is even #(-1)^n=color(red)(-1)#
#-------------------#
#color(blue)(x=(-1)^nA+npi)#
where #A=sin^-1(a)#
#ninNN#
