Trigonometric identities and equations 155
1.0
21.0
20.6
0
908
2708
323.138
216.878
y 5 sin ␪
T
S
A
C
908
1808
2708
3608
08
a
a
(a)
(b)
y
␪
1808
3608
Figure 15.3
Tangent is positive in the first and third quadrants (see
Fig. 15.4).
The acute angle tan −1 1.2000 =50.19 ◦ . Hence,
x = 50.19
◦ or 180
◦
+ 50.19
◦
= 230.19
◦
(a)
(b)
50.198
50.198
908
C
A
S
T
2708
1808
3608
08
y
x
1.2
0
50.198
y 5 tan x
230.198
908
1808
2708
3608
Figure 15.4
Problem 8. Solve for θ in the range
0 ◦ ≤ θ ≤ 360 ◦ for 2 sin θ = cos θ
Dividing both sides by cos θ gives:
2 sin θ
cos θ
= 1
From Section 15.1, tan θ =
sin θ
cos θ
,
hence 2 tan θ = 1
Dividing by 2 gives: tan θ =
1
2
from which, θ = tan −1 1
2
Since tangent is positive in the first and third quadrants,
θ = 26.57
◦ and 206.57
◦
Problem 9. Solve 4 sec t = 5 for values of t
between 0 ◦ and 360 ◦ .
4 sect = 5, from which sec t =
5
4 = 1.2500
Hence t = sec −1 1.2500
Secant = (1/cosine) is positive in the first and
fourth quadrants (see Fig. 15.5) The acute angle
sec −1 1.2500 =36.87 ◦ . Hence,
t = 36.87
◦ or 360
◦
− 36.87
◦
= 323.13
◦
36.878
36.878
908
C
A
S
T
2708
1808
3608
08
Figure 15.5
Now try the following exercise
Exercise 66 Further problems on
trigonometric equations
In Problems 1 to 3 solve the equations for angles
between 0 ◦ and 360 ◦ .
1. 4 −7 sinθ = 0
[ θ = 34.85 ◦ or 145.15 ◦ ]
2. 3 cosec A + 5.5 =0
[A = 213.06
◦ or 326.94
◦ ]
3. 4(2.32 − 5.4 cot t ) = 0
[t = 66.75 ◦ or 246.75 ◦ ]