This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ctan(−3). Since −3 < 0, − π < arctan(−3) < 0, so multiplying through by 2 tells us
−π < 2 arctan(−3) < 0 which is not in the range [0, 2π ). Hence, we discard this answer along
with all other answers obtained for k < 0. Starting with the positive integers, for k = 1 we
ﬁnd x = 2 arctan(−3) + 2π . Since −π < 2 arctan(−3) < 0, we get that x = 2 arctan(−3) + 2π
is between π and 2π , so we keep this solution. For k = 2, we get x = 2 arctan(−3) + 4π , and
The reader is encouraged to see what happens if we had chosen the reciprocal identity cot(3x) = tan(3x) instead.
The graph on the calculator appears identical, but what happens when you try to ﬁnd the intersection points?
Geometrically, we are ﬁnding the measures of all angles with a reference angle of π . Once again, visualizing these
numbers as angles in radian measure can help us literally ‘see’ how these two families of solutions are related. 732 Foundations of Trigonometry
since 2 arc...
View Full Document