Unizor - Trigonometry - Inequalities
Channel: Zor Shekhtman on UNIZOR Education
Date: 2013-11-18

As a general recommendation on how to solve any inequality of a type f(x) ≤ a, consider a corresponding equation f(x) = a and then examine the value of a function y = f(x) in between the solutions to this equation. Are they greater or less than a? Those intervals where the values are less are the solutions (including the endpoints). Alternatively, we can examine the sign of a function y = f(x) − a on any interval in between the solutions to this equation.

Inequality tan(x) ≤ a

Consider a graph of a function y=tan(x). This is a curve that has a period of π and on a base interval from −π/2 to π/2 (not including the ends where it's not defined) it changes from negative infinity through point (0,0) to positive infinity. Then it repeats this behavior periodically throughout all the real values of an argument x.
On the step 1 of solving our equation we will find its solution in the base interval. On this base interval a function y=tan(x) is monotonic and increasing, taking all real values as an argument moves from −π/2 to π/2. Therefore, for any real a there will be a contiguous interval from −π/2 (not including) to a solution of the equation tan(x)=a (including) where tan(x) is less or equal to a. The solution of the equation tan(x)=a on a base interval of periodicity exists, is unique and equals to x=arctan(a).
Therefore, a solution of an inequality tan(x) ≤ a on an interval from −π/2 to π/2 is −π/2 LESS THAN x LESS OR EQUAL TO arctan(a).
On the step 2 we will expand our base interval solution to all values of an argument using periodicity. 

Inequality cot(x) ≤ a

Inequalities involved cotangent are completely similar to those of tangent. The only difference is that cotangent is a monotonically decreasing function and its base interval of periodicity is between the points 0 and π (not including the endpoints where cotangent is undefined). Therefore, the solution of the inequality cot(x) ≤ a on a base interval of periodicity lies to the right of the point x=arccot(a). The complete solution to our inequality is obtained by expansion using the periodicity.

Inequality sin(x) ≤ a

This is slightly more complicated. Relative simplicity of solving trigonometric inequalities tan(x) ≤ a and cot(x) ≤ a was based on a fact that within a base interval of periodicity functions tangent and cotangent behaved monotonically and inverse functions arctangent and arccotangent were defined exactly on the same base interval. Let's consider an interval of periodicity from −π to π. It's chosen because it includes the interval from −π/2 to π/2 where inverse function arcsine is defined. The function sine is not monotonic on the chosen interval of periodicity, so the same approach as with tangent would not work so easily.

We can use the graph of a function y=sin(x) to find the solutions as we did above or when considering the equation sin(x)=a or, alternatively, view a sine as an ordinate of a point on a unit circle. Then, to find a solution to an inequality sin(x) ≤ a, we have to find all central angles whose corresponding points on a unit circle lie below or on a horizontal line y=a.

For a greater than 1 the solution set includes all the possible angles since the line y=a lies above the unit circle.
For a less than −1 this set is empty since the line y=a lies below the unit circle.
Non-trivial solutions are in case
−1 ≤ a ≤ 1.
In case 0 ≤ a ≤ 1 (and, therefore, 0 ≤ arcsin(a) ≤ π/2) the central angles on the unit circle (from −π to π) whose corresponding points have ordinates less than or equal to a are those from −π to arcsin(a) and from π−arcsin(a) to π. Thus, it's two intervals (including both endpoints in both cases).
In case −1 ≤ a ≤ 0 (and, therefore, −π/2 ≤ arcsin(a) ≤ 0) the central angles on the unit circle (from −π to π) whose corresponding points have ordinates less than or equal to a are those from −π−arcsin(a) to arcsin(a). Thus, it's one consecutive interval (including endpoints).

Adding 2π·N to all the above solutions completes the specification of all the solutions to our inequality.

Inequality cos(x) ≤ a

Similarly to the previous case with sine, consider a unit circle where cosine is an abscissa of a central angle. Let's choose an interval of periodicity for cosine to be from −π to π, which includes the interval from 0 to π where an inverse function arccosine is defined.

As in the case with sine, the solution to our inequality when a is greater than 1 are all the real values of an argument and, if a is less than −1, there are no solutions.

For any a between −1 and 1 solutions within an interval of periodicity lie in two areas: from −π to −arccos(a) and from arccos(a) to π (including both endpoints in both cases).

Adding 2π·N to all the above solutions completes the specification of all the solutions to our inequality.