Note that the denominator is undefined whenever

#4 sin(x) + 2 = 0#,

that is, whenever

#x = x_(1,n) = pi/6 + n 2pi#

or

#x = x_(2,n) = (5 pi)/6 + n 2pi#,

where #n in ZZ# (#n# is an integer).

As #x# approaches #x_(1,n)# from below, #f(x)# approaches #- infty#, while if #x# approaches #x_(1,n)# from above then #f(x)# approaches #+infty#. This is due to division by "almost #-0# or #+0#".

For #x_(2,n)# the situation is reversed. As #x# approaches #x_(2,n)# from below, #f(x)# approaches #+infty#, while if #x# approaches #x_(2,n)# from above then #f(x)# approaches #-infty#.

We get a sequence of intervals in which #f(x)# is continuous, as can be seen in the plot. Consider first the "bowls" (at whose ends the function blows up to #+infty#). If we can find the local minima in these intervals, then we know that #f(x)# assumes all the values between this value and #+infty#. We can do the same for "upside-down bowls", or "caps".

We note that the smallest positive value is obtained whenever the denominator in #f(x)# is as large as possible, that is when #sin(x) = 1#. So we conclude that the smallest positive value of #f(x)# is #1/(4*1 + 2) = 1/6#.

The largest negative value is similarly found to be #1/(4*(-1) + 2) = -1/2#.

Due to the continuity of #f(x)# in the intervals between discontinuities, and the Intermediate value theorem , we can conclude that the range of #f(x)# is

#R = (-infty, -1/2] uu [1/6, +infty)#

The hard brackets mean that the number is included in the interval (e.g. #-1/2#), while soft brackets means that the number is not included.

graph{1/(4sin(x) + 2) [-10, 10, -5, 5]}