At inequalitiestrigonometric are inequalities that have at least one trigonometric ratio wherein angle is unknown. the unknown of a inequalitytrigonometric it is a bow, therefore, just as in inequalities the solution is given by an interval, in trigonometric inequalities, too. The difference is that this interval is an arc in the trigonometric cycle, in which each point corresponds to an angle that can be considered the result of the inequality.
In this article, we'll solve the inequalityfundamentalsenx> k. The solution of this inequality is analogous to the solution of the inequalities senx < k, senx ≤ k and senx ≥ k.
Trigonometric cycle and the solution of the inequality
The solutions of inequalitysenx > k they are in cycletrigonometric. Therefore, k must be in the range [–1, 1]. This interval is on the y axis of the Cartesian plane, which is the sine axis. The interval in which the value of x is located is an arc of the trigonometric cycle.
Assuming that k is in the interval [0, 1], we have the following image:
In the axis of sines (y axis), the values that cause senx > k are those above point k. The arc that includes all these values is the smallest, DE, illustrated in the figure above.
The solution of inequalitysenx > k considers all values of x (which is an angle) between point D and point E of the cycle. Assuming that the smallest arc BD is related to angle α, this means that the angle related to the smallest arc, BE, measures π – α. So, one of the solutions to this problem is the interval that goes from α to π – α.
This solution is only valid for the first round. If there is no restriction for the inequalitytrigonometric, we must add the portion 2kπ, which indicates that k turns can be done.
Therefore, the algebraic solution of inequalitysenx> k, when k is between 0 and 1, it is:
S = {xER| α + 2kπ < x < π – α + 2kπ}
With k belonging to natural set.
Note that for the first round, k = 0. For the second round, we have two results: the first, where k = 0, and the second, where k = 1. For the third round, we will have three results: k = 0, k = 1 and k = 2; and so on.
In which case k is negative
When k is negative, the solution can be obtained in the same way as explained above. So, we will have in the cycletrigonometric:
The difference between this case and the previous one is that, now, the angle α is related to the larger arc BE. So the measure of this arc is π + α. The largest arc BD measures 2π – α. So, the solutiongivesinequalitysenx > k, for negative k, is:
S = {xER| 2π – α + 2kπ < x < π + α + 2kπ}
Furthermore, the 2kπ portion appears in this solution for the same reason mentioned before, related to the number of turns.
by Luiz Moreira
Graduated in Mathematics
Source: Brazil School - https://brasilescola.uol.com.br/matematica/solucao-inequacao-fundamental-senx.htm