We will here solve the inequality by dividing by x, and we have to distinguish two cases:

a) x > 0. Then we just divide and we get sin(2x + 1) > 0. This can be solved for instance by imagining the graph of sin(y).

We see that sin(y) is positive exactly if y is from the interval (0,π) and its shifts by 2kπ for k any integer. For our case this means that the expression 2x + 1 must belong to one of the intervals (2kπ,π + 2kπ). We can write it as an inequality (or rather, two inequalities) 2kπ < 2x + 1 < π + 2kπ, solving it for x (we can solve both simultaneously) we get

where k can be any integer.

However, now we have to recall that we were doing this solution only under the assumption that x > 0, so we can only take the part of this solution that includes positive numbers. If k > 1; the corresponding intervals lie on the positive x-axis and so they survive. For k < 0, the corresponding intervals only include negative numbers and so they can be ignored. If k = 0, we get the interval (−1/2,π/2 − 1/2) whose part lies in positive numbers, so we take only that part. Thus we get

b) x < 0. Now when we divide we have to switch the direction of the inequality, thus we get sin(2x + 1) < 0. As before, from the graph of the sine function we conclude that now 2x + 1 must belong to one of the intervals (−π + 2kπ,2kπ). We can think of it as of inequality -π + 2kπ < 2x + 1 < 2kπ, solving it we get

where k can be any integer.

As before, we have to recall now that we were doing this solution only under the assumption that x < 0, so we can only take the part of this solution that includes negative numbers. Thus we get

c) To cover all possibilities, we should actually also check what happens when x = 0, and we immediately see that this is not a solution of the given inequality, so this case does not add to our solution.

Since we had a choice of three variants here, we get the final solution by taking a union of the partial solutions: