Question 710041
{{{tan(x)-sqrt(3)<=0}}}
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 {{{2pi}}}.<br>
Adding the square root to each side:
{{{tan(x)<=sqrt(3)}}}
{{{sqrt(3)}}} is a special angle value for tan. The reference angle is {{{pi/3}}}. Since the period of tan is just {{{pi}}}, the set of angles that makes {{{tan(x)=sqrt(3)}}} would be:
{{{x = pi/3 + pi*n}}} (where "n" is any integer)<br>
For the "less than" part of the inequality we should realize:<ul><li>All negative tan's will be less than a positive number like {{{sqrt(3)}}}. Since tan is negative in the 2nd and 4th quadrants, <i>all</i> angles which terminate in either of these quadrants will be solutions to this inequality.</li><li>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 {{{sqrt(3)}}} we want reference angles that are less than {{{pi/3}}}. IOW: We want a reference angle between 0 and {{{pi/3}}} in quadrants where tan is positive (1st and 3rd).</li><li>If tan(x) is zero then it would be less than {{{sqrt(3)}}}. So x's that make tan zero would also be solutions to the inequality:
{{{x = 0 + pi*n}}}</li></ul>To summarize, the solutions to your inequality are:<ul><li>{{{x = pi/3 + pi*n}}} (because they make tan(x) equal to {{{sqrt(3)}}}</li><li>{{{x = 0 + pi*n}}} (because they make tan(x) equal to 0 (which is less than {{{sqrt(3)}}})</li><li>x = any angle that terminates in the 1st or 3rd quadrant with a reference angle less than {{{pi/3}}} (because they make tan(x) a positive number less than {{{sqrt(3)}}}</li><li>x = all angles that terminate in the 2nd or 4th quadrant (because they make tan(x) negative (which is less than {{{sqrt(3)}}}))</li></ul>