You can put this solution on YOUR website!
Is this the whole problem? I suspect that there may be something about finding solutions from within some set of numbers, like between 0 and .
Adding the square root to each side: is a special angle value for tan. The reference angle is . Since the period of tan is just , the set of angles that makes would be: (where "n" is any integer)
For the "less than" part of the inequality we should realize:
All negative tan's will be less than a positive number like . Since tan is negative in the 2nd and 4th quadrants, all angles which terminate in either of these quadrants will be solutions to this inequality.
For positive tan's, the tan value grows as the reference angle grows and vice versa. So for positive tan's that are less than we want reference angles that are less than . IOW: We want a reference angle between 0 and in quadrants where tan is positive (1st and 3rd).
If tan(x) is zero then it would be less than . So x's that make tan zero would also be solutions to the inequality:
To summarize, the solutions to your inequality are:
(because they make tan(x) equal to
(because they make tan(x) equal to 0 (which is less than )
x = any angle that terminates in the 1st or 3rd quadrant with a reference angle less than (because they make tan(x) a positive number less than
x = all angles that terminate in the 2nd or 4th quadrant (because they make tan(x) negative (which is less than ))
(