Determination of base or fundamental interval is central to solve trigonometric inequality. The function values in this interval is repeated with a periodicity of trigonometric function. The base interval depends on the nature of trigonometric function and inequality in question. The steps to find solution of trigonometric inequality are :

1 : Convert given inequality to trigonometric equation by replacing inequality sign by equality sign.

2 : Solve resulting equation in the interval [0,2π]. There are two solutions. They are the angle values at which trigonometric function has the value which is being compared in the given inequality.

3 : Convert positive angle greater than π to equivalent negative value to account for the fact that basic interval being repeated may lie on negative side of the origin (cosine, secant and tangent function).

4 : Construct base interval between two values, keeping in mind the given inequality. It is always advantageous to draw a rough intersection of graphs of each side of given inequality.

5 : If function asymptotes (tangent, cotangent, secant and cosecant) within the interval constructed, then basic interval is limited by the angle value at which function asymptotes.

6 : Generalize solution by extending base interval with the period of the trigonometric function.

In order to understand the process, let us solve the inequality given by :

This example has been selected here as it involves consideration of each step as enumerated above for finding solution of inequality. Corresponding trigonometric equation, in this case, is :

The acute angle is π/3. Further, tangent function is negative in second and fourth quarter (see sign diagram). Using value diagram in conjunction with sign diagram, solution of given equation in [0, 2π] are :

Sign and value diagram |
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Here, second angle is greater than π. Hence, equivalent negative angle is :

Tangent function, however, is not a continuous function between -π/3 and 2π/3. Tangent values are greater than -√3 for angle greater than -π/3, but value asymptotes to infinity at π/2. This can be verified from the intersection graph.

Trigonometric inequality |
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Thus, basic interval satisfying inequality is :

It is also clear that the solution in this interval is repeated with a period of π, which is period of tangent function. Hence, solution of given inequality is :