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Evaluating trigonometric ratios is a direct process in which we make use of known values, trigonometric identities and transformations or even pre-defined trigonometric tables. The evaluation of trigonometric inequalities is somewhat inverse of this process. Consider an inequality :
$$\mathrm{tan}x\ge -\sqrt{3}$$
Clearly, we need to know “x” for which this inequality holds. As pointed out earlier, trigonometric functions are “many-one” relation. The value of “x” satisfying given inequality is not an unique interval, but a series of intervals. Incidentally, however, trigonometric function values repeat after certain “period”. So this enables us to define periodic intervals in generic manner for which trigonometric inequality holds.
We observe that line $y=-\sqrt{3}$ intersects tangent graph at multiple points. The sections of plots satisfying the inequality are easily identified on the graph and are shown as dark red line.
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 :
$$\mathrm{tan}x\ge -\sqrt{3}$$
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 :
$$\mathrm{tan}x=-\sqrt{3}$$
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 :
$$\Rightarrow x=\pi -\theta =\pi -\frac{\pi}{3}=\frac{2\pi}{3}$$ $$\Rightarrow x=2\pi -\theta =2\pi -\frac{\pi}{3}=\frac{5\pi}{3}$$
Here, second angle is greater than π. Hence, equivalent negative angle is :
$$\Rightarrow y=\frac{5\pi}{3}-2\pi =-\frac{\pi}{3}$$
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