082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

3.8 Trigonometric equations 147
tan t
2
1.3100
1
0
p
A –2 3p
B p C –2 D 2p t
–1
–1.3100
–2
Figure3.22
Aand Care solution pointsfortant = 1.3100. B andDare
solution pointsfortant = −1.3100.
This isthe solution between 0 and π given by point A.UsingFigure 3.22 wecan
2
see thatthe second solution isgiven by
t = π +0.9188 = 4.0604
This isgiven by point C.
(b) Figure 3.22 shows that there are two solutions of tant = −1.3100, one between π 2
and π,theotherbetween 3π 2 and2π.PointsBandDrepresentthesesolutions.Using
a scientific calculator we have
t = tan −1 (−1.3100) = −0.9188
This solution is outside the range of interest. Noting that the period of tant is π we
see that
t = −0.9188 + π = 2.2228
isasolution between π and π. This isgiven by pointB. The second solution is
2
t = −0.9188 +2π = 5.3644
This is the solution given by point D. The required solutions aret = 2.2228 and
5.3644.