Find ALL the solutions to the following then draw them on the unit circle:
tan(x) = -3.4076 on the interval [0,2Pi)
2007-07-22 12:31:13 · 3 answers · asked by francis r 1 in Science & Mathematics ➔ Mathematics
Wait..I am now a bit confused because isn't 106 degrees in Quad II and 286 degrees in Quad IV?
I feel really stupid, sorry..
2007-07-22 13:36:03 · update #1
tg(x) = -3.4076
x= -1.285 rad = -73º 38' 42.21 '' ( arctg(-3.4076 rad) )
in the trigonometric circle you draw 2 solutions:
one at -73 º ... (that's in the 4th quadrant)
and the symmetrical in the 2th quadrant (180º - 73º)
The total solutions (look at the graph of tangent) for [-2pi;2pi] are 4:
x= -73º ; x= -90º-73º; x=180º-73º; x=360º-73º
The unitary circle refers to the positive (0, 2pi) part of the cartesian graph but the tangent goes -8;+8 so only 2 solutions are found in the unitary circle (0; 2pi) but the function has 4 in (-2pi; 2pi).
2007-07-22 15:11:37 · answer #1 · answered by GPC 3 · 0⤊ 0⤋
Draw the circle.
arctan(-3.4076) is an angle in Quadrant III.
An angle of {[arctan(-3.4076)] - pi} also has a tangent of -3.4076.
2007-07-22 20:28:31 · answer #2 · answered by Mark 6 · 0⤊ 0⤋
x = arctan(-3.4076) for x in the interval [0, 2Ï).
x â 106.35494° and 286.35494°
2007-07-22 19:43:08 · answer #3 · answered by Northstar 7 · 0⤊ 0⤋