Trigonometric identities and equations 155
1.02 1.02 0.6(^09082708)
216.87 8 323.13 8
y 5 sin
T
S A
C
908
1808
2708
3608
08
a a
(a)
(b)
y
(^18083608)
Figure 15.3
Tangent is positive in the first and third quadrants (see
Fig. 15.4).
The acute angle tan−^11. 2000 = 50. 19 ◦. Hence,
x= 50. 19 ◦or 180◦+ 50. 19 ◦= 230. 19 ◦
(a)
(b)
50.19 8
50.19 8
908
C
S A
T
2708
1808
3608
08
y
x
1.2
0
50.19 8
y 5 tan x
230.19 8
908 1808 2708 3608
Figure 15.4
Problem 8. Solve forθin the range
0 ◦≤θ≤ 360 ◦for 2sinθ=cosθ
Dividing both sides by cosθgives:
2sinθ
cosθ
= 1
From Section 15.1, tanθ=
sinθ
cosθ
,
hence 2tanθ= 1
Dividing by 2 gives: tanθ=^12
from which,θ=tan−^112
Since tangent is positive in the first and third quadrants,
θ= 26. 57 ◦and 206. 57 ◦
Problem 9. Solve 4sect=5 for values oft
between 0◦and 360◦.
4sect=5, from which sect=^54 = 1. 2500
Hencet=sec−^1 1.2500
Secant=( 1 /cosine) is positive in the first and
fourth quadrants (see Fig. 15.5) The acute angle
sec−^11. 2500 = 36. 87 ◦. Hence,
t= 36. 87 ◦or 360◦− 36. 87 ◦= 323. 13 ◦
36.87 8
36.87 8
908
C
S A
T
2708
1808
3608
08
Figure 15.5
Now try the following exercise
Exercise 66 Further problemson
trigonometric equations
In Problems 1 to 3 solve the equations for angles
between 0◦and 360◦.
- 4−7sinθ=0[θ= 34. 85 ◦or 145. 15 ◦]
- 3cosecA+ 5. 5 = 0
[A= 213. 06 ◦or 326. 94 ◦] - 4( 2. 32 − 5 .4cott)= 0
[t= 66. 75 ◦or 246. 75 ◦]