Vertical symptotes occur where f(x) is undefined. This will occur where the function is over 0 in this case. We must therefore determine when the denominator equals 0. Whenever two things are multiplied together, such as in this case 3x-2 and sin(pi+(2*pi)/x), the total will be equal to zero when either or both are zero so we must find when both equal 0. To do this, set each expression separately equal to 0 and solve for x.
Now we have the equations 3x-2=0 and sin(pi+(2*pi)/x)=0. The former is much easier to solve. Simply add two to both sides and divide both sides by three to get: x=2/3.
For the next equation we must first find where sine is equal to 0. This only occurs at the quadrantals of 0, pi, and 2*pi within a range of 0 to 2*pi for the angle. Since 2*pi and 0 are the same angle we can disregard one. This means that sine is equal to 0 whenever its argument is 0 or pi. From this we can make two MORE equations from setting the argument equal to these two things.
This gives us: pi+(2*pi)/x=0 and pi+(2*pi)/x=pi. For the second equation subtract pi from both sides and multiply both sides by x to get: 2*pi=0 which means that this case can never happen because 2pi will NEVER equal 0 and we can disregard it. For the first equation factor out 2pi from the left expression to get 2pi*(1/2+1/x)=0 then divide by 2pi on both sides to get 1/2+1/x=0. Subtract 1/2 from both sides and multiply by x on both sides to get 1=-x/2 then multiply by -2 on both sides to finally get: x=-2.
Finally if we combine all of the x's we found we get that there are vertical asymptotes for this function at x={-2,2/3} within a range of 0 to 2pi. Hope I helped!
