Answer:
0, 4, -4 and they may want you to mention formally all the kt multiples of [tex]\pi[/tex].
Step-by-step explanation:
Let's do the second derivative of the function: [tex]y(t)=sin(k\,t)[/tex]
[tex]y'(t) =k\,cos(k\,t)\\y"(t)=-k^2\,sin(kt)[/tex]
So now we want:
[tex]y"+16\,y'=0\\-k^2\,sin(kt)+\,16\,sin(kt)=0\\sin(kt)\,(16-k^2)=0\\[/tex]
Then we have to include the zeros of the binomial ([tex]16-k^2[/tex]) which as you say are +4 and -4, and also the zeros of [tex]sin(kt)[/tex], which include all those values of
[tex]kt=0\,,\,\pi\,\,,\,2\pi\, ,\,etc.[/tex]
So an extra one that they may want you to include is k = 0
