The way I would approach this is the following:
Let f(x) = x^{179} + 163/(1+x^2+\sin^2(x)) - 119. In order to show what you wanted to show, it suffices to show that there exists some x_0 such that f(x_0) = 0. We can do that by showing that f is positive somewhere and negative somewhere else; then if f is continuous, the intermediate value theorem says that f has a zero somewhere in the middle. The thing that pops out to check about continuity in this is the denominator of that rational expression: 1+x^2+\sin^2(x). We need to make sure that it is never zero. I'll leave that part to you, but it's pretty simple when you consider what would need to happen in order for it to be zero.
Now we just need to find a positive and a negative number for f. If x << 0 (i.e. hugely negative), then x^{179} is hugely negative. The rational expression is positive and small, and the -119 would certainly nullify it. Therefore for very negative x, f is negative (in fact, x = -2 would work). The same logic applies for x >> 0 (or even x = 2). So we have a continuous function which is negative on one part of its domain and positive on the other; therefore, it must be zero in between.
