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

3.8 Trigonometric equations 145
sin t
1
0.6105
0 p –2 3p –2
Figure3.20
A B p C D 2p
t
–0.6105
–1
Aand Bare solution pointsfor
sint = 0.6105.Cand Dare
solution pointsforsint = −0.6105.
and so, using a calculator, wesee
t = sin −1 (−0.6105) = −0.6567
Although this value oft is a solution of sint = −0.6105 it is outside the range of
values of interest.Recall that
sint = sin(t +2π)
that is,adding 2πtoanangle does notchange the sineof the angle. Hence
t = −0.6567 +2π = 5.6265
is a required solution. This is the solution given by point D. From the symmetry of
Figure 3.20 the other solution is
t = π +0.6567 = 3.7983
This isthe solution atpoint C. The required solutions aret = 3.7983, 5.6265.
Example3.16 Solve
(a) cost = 0.3685 for0 t 2π
(b) cost = −0.3685 for0 t 2π
Solution Figure 3.21 shows a graph of y = cost between t = 0 and t = 2π together with
horizontallines aty = 0.3685andy = −0.3685.
(a) FromFigure3.21weseethatthereisasolutionofcost = 0.3685between0and π 2
and a solution between 3π 2
and 2π. These aregiven by points Aand D.Now
cost = 0.3685
and so, using a scientific calculator, wesee
t = cos −1 (0.3685) = 1.1934