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

Usingascientific calculator wehave
z = sin −1 (0.5) = 0.5236
3.8 Trigonometric equations 149
This solution is outside the range of interest. By reference to Figure 3.7 the next solution
is
z = π −0.5236 = 2.6180
This isthe first value ofzgreater than 1 such thatsinz = 0.5. Finally
t=z−1=1.6180
The voltage first has a value of 1.5 volts whent = 1.618 seconds.
EXERCISES3.8
1 Solvethe followingequations for0 t 2π:
(a) sint = 0.8426 (b) sint = 0.2146
(c) sint = 0.5681 (d) sint = −0.4316
(e) sint = −0.9042 (f) sint = −0.2491
2 Solvethe followingequations for0 t 2π:
(a) cost = 0.4243 (b) cost = 0.8040
(c) cost = 0.3500 (d) cost = −0.5618
(e) cost = −0.7423 (f) cost = −0.3658
3 Solvethe followingequations for0 t 2π:
(a) tant = 0.8493 (b) tant = 1.5326
(c) tant = 1.2500 (d) tant = −0.8437
(e) tant = −2.0612 (f) tant = −1.5731
4 Solvethe followingequations for0 t 2π:
(a)sin2t = 0.6347
(b)sin3t = −0.2516
( )
t
(c)sin = 0.4250
2
(d)sin(2t +1) = −0.6230
(e)sin(2t −3) = 0.1684
( )
t +2
(f) sin = −0.4681
3
5 Solve the following equations for0 t 2π:
(a) cos2t = 0.4234
( )
t
(b) cos = −0.5618
3
( )
2t
(c) cos = 0.6214
3
(d) cos(2t +0.5) = −0.8300
(e) cos(t −2) = 0.7431
(f) cos(πt −1) = −0.5325
6 Solve the following equations for0 t 2π:
(a) tan2t = 1.5234
( )
t
(b) tan = −0.8439
3
(c) tan(3t −2) = 1.0641
(d) tan(1.5t −1) = −1.7300
( )
2t+1
(e) tan = 1.0000
3
(f) tan(5t −6) = −1.2323
7 A time-varying voltage, v(t),has the form
v(t) =20sin(50πt +20)
t 0
Calculate the firsttime that the voltage has avalue of
(a)2volts (b)10volts (c)15volts
Solutions
1 (a) 1.0021,2.1395 (b) 0.2163,2.9253
(c) 0.6042,2.5374 (d) 3.5879,5.8369
(e) 4.2711,5.1537 (f) 3.3933,6.0314
2 (a) 1.1326,5.1506 (b) 0.6368,5.6464
(c) 1.2132,5.0700 (d) 2.1674,4.1158
(e) 2.4073,3.8759 (f) 1.9453,4.3379