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

148 Chapter 3 The trigonometric functions
Example3.18 Solve
(a) sin2t = 0.6105 for0t 2π
(b) cos(3t +2) = −0.3685for0 t 2π
Solution (a) Letz = 2t.Ast variesfrom0to2πthenzvariesfrom0to4π.Thustheproblemis
equivalent tosolving
sinz=0.6105
0z4π
From Example 3.15 the solutions between 0 and 2π are 0.6567 and 2.4849. Since
sinz has period 2π, then the solutions in the next cycle, that is between 2π and 4π,
arez = 0.6567 +2π = 6.9399 andz = 2.4849 +2π = 8.7681. Hence
z = 2t = 0.6567,2.4849,6.9399,8.7681
and so, tofour decimal places,
t = 0.3284,1.2425,3.4700,4.3840
(b) Letz = 3t +2. Ast varies from 0 to 2π thenzvaries from 2 to 6π +2. Hence the
problem isequivalent tosolving
cosz=−0.3685
2z6π+2
Solutions between 0 and 2π are given in Example 3.16 as z = 1.9482, 4.3350.
Notingthatcoszhasperiod2π,thensolutionsbetween2πand4πarez = 1.9482+
2π = 8.2314 andz = 4.3350 + 2π = 10.6182, solutions between 4π and 6π are
z = 1.9482+4π = 14.5146andz = 4.3350+4π = 16.9014andsolutionsbetween
6π and 8π arez = 1.9482 +6π = 20.7978 andz = 4.3350 +6π = 23.1846. The
solutions betweenz = 0 andz = 8πarethus
z = 1.9482,4.3350,8.2314,10.6182,14.5146,16.9014,20.7978,23.1846
Noting that 6π + 2 = 20.8496 we require values of z between 2 and 20.8496,
thatis
Finally
z = 4.3350,8.2314,10.6182,14.5146,16.9014,20.7978
t = z −2
3
= 0.7783,2.0771,2.8727,4.1715,4.9671,6.2659
Example3.19 Avoltage, v(t), isgiven by
v(t)=3sin(t+1)
t 0
Findthe first time thatthe voltage has a value of1.5 volts.
Solution We needtosolve 3sin(t +1) = 1.5,thatis
sin(t+1)=0.5
t0
Letz =t +1.Sincet 0 thenz 1.Theproblemisthus equivalent to
sinz=0.5
z1