Trigonometric identities and equations 157
15.6 Worked problems (iii) on
trigonometric equations
Problem 12. Solve the equation
8 sin
2
θ + 2 sin θ − 1 = 0,
for all values of θ between 0 ◦ and 360 ◦ .
Factorizing 8 sin
2
θ + 2 sinθ − 1 = 0 gives
(4 sin θ − 1) (2 sin θ + 1) = 0.
Hence 4 sin θ − 1 = 0, from which, sin θ =
1
4 = 0.2500,
or 2 sin θ + 1 =0, from which, sin θ =−
1
2 = −0.5000.
(Instead of factorizing, the quadratic formula can, of
course, be used).
θ = sin
−1 0.2500 = 14.48 ◦ or 165.52 ◦ , since sine
is positive in the first and second quadrants, or
θ = sin −1 (−0.5000) = 210 ◦ or 330 ◦ , since sine is negative in the third and fourth quadrants. Hence
θ = 14.48
◦
, 165.52
◦
, 210
◦ or 330
◦
Problem 13. Solve 6 cos 2 θ + 5 cosθ − 6 = 0 for
values of θ from 0 ◦ to 360 ◦ .
Factorizing 6 cos 2 θ + 5 cos θ − 6 =0 gives
(3 cos θ − 2) (2 cos θ + 3) = 0.
Hence 3 cosθ − 2 = 0, from which, cos θ =
2
3 = 0.6667,
or 2 cos θ + 3 =0, from which, cos θ = −
3
2 =−1.5000.
The minimum value of a cosine is −1, hence the latter expression has no solution and is thus neglected.
Hence,
θ = cos
−1 0.6667 = 48.18
◦ or 311.82
◦
since cosine is positive in the first and fourth quadrants.
Now try the following exercise
Exercise 68 Further problems on
trigonometric equations
In Problems 1 to 3 solve the equations for angles
between 0 ◦ and 360 ◦ .
1. 15 sin 2 A + sin A − 2 = 0
A = 19.47 ◦ , 160.53 ◦ ,
203.58 ◦ or 336.42 ◦
2. 8 tan 2 θ + 2 tan θ = 15 θ = 51.34 ◦ , 123.69 ◦ ,
231.34 ◦ or 303.69 ◦
3. 2 cosec 2 t − 5 cosec t = 12
t = 14.48 ◦ , 165.52 ◦ ,
221.81 ◦ or 318.19 ◦
4. 2 cos 2 θ + 9 cosθ − 5 = 0
[θ = 60 ◦ or 300 ◦ ]
15.7 Worked problems (iv) on
trigonometric equations
Problem 14. Solve 5 cos 2 t + 3 sint − 3 =0 for
values of t from 0 ◦ to 360 ◦ .
Since cos 2 t + sin 2 t = 1, cos 2 t = 1 − sin 2 t . Substituting
for cos 2 t in 5 cos 2 t + 3 sin t − 3 = 0 gives:
5(1 − sin
2 t ) + 3 sint − 3 = 0
5 − 5 sin
2 t + 3 sint − 3 = 0
−5 sin
2 t + 3 sint + 2 = 0
5 sin
2 t − 3 sint − 2 = 0
Factorizing gives (5 sin t + 2)(sin t − 1) = 0. Hence
5 sint + 2 = 0, from which, sint =−
2
5 =−0.4000, or
sin t − 1 =0, from which, sin t = 1.
t = sin −1 (−0.4000) = 203.58 ◦ or 336.42 ◦ , since sine
is negative in the third and fourth quadrants, or
t = sin −1 1 = 90 ◦ . Hence t = 90 ◦ , 203.58
◦ or 336.42 ◦
as shown in Fig. 15.7.
1.0
y
t 8
20.4
21.0
0
9 0 8
203.588
y 5 sin t
336.428
2708
3608
Figure 15.7
Problem 15. Solve 18 sec 2 A − 3 tan A = 21 for
values of A between 0 ◦ and 360 ◦ .
1 + tan 2 A = sec 2 A. Substituting for sec 2 A in
18 sec 2 A − 3 tan A = 21 gives
18(1 + tan 2 A) − 3 tan A = 21,
15.6 Worked problems (iii) on
trigonometric equations
Problem 12. Solve the equation
8 sin
2
θ + 2 sin θ − 1 = 0,
for all values of θ between 0 ◦ and 360 ◦ .
Factorizing 8 sin
2
θ + 2 sinθ − 1 = 0 gives
(4 sin θ − 1) (2 sin θ + 1) = 0.
Hence 4 sin θ − 1 = 0, from which, sin θ =
1
4 = 0.2500,
or 2 sin θ + 1 =0, from which, sin θ =−
1
2 = −0.5000.
(Instead of factorizing, the quadratic formula can, of
course, be used).
θ = sin
−1 0.2500 = 14.48 ◦ or 165.52 ◦ , since sine
is positive in the first and second quadrants, or
θ = sin −1 (−0.5000) = 210 ◦ or 330 ◦ , since sine is negative in the third and fourth quadrants. Hence
θ = 14.48
◦
, 165.52
◦
, 210
◦ or 330
◦
Problem 13. Solve 6 cos 2 θ + 5 cosθ − 6 = 0 for
values of θ from 0 ◦ to 360 ◦ .
Factorizing 6 cos 2 θ + 5 cos θ − 6 =0 gives
(3 cos θ − 2) (2 cos θ + 3) = 0.
Hence 3 cosθ − 2 = 0, from which, cos θ =
2
3 = 0.6667,
or 2 cos θ + 3 =0, from which, cos θ = −
3
2 =−1.5000.
The minimum value of a cosine is −1, hence the latter expression has no solution and is thus neglected.
Hence,
θ = cos
−1 0.6667 = 48.18
◦ or 311.82
◦
since cosine is positive in the first and fourth quadrants.
Now try the following exercise
Exercise 68 Further problems on
trigonometric equations
In Problems 1 to 3 solve the equations for angles
between 0 ◦ and 360 ◦ .
1. 15 sin 2 A + sin A − 2 = 0
A = 19.47 ◦ , 160.53 ◦ ,
203.58 ◦ or 336.42 ◦
2. 8 tan 2 θ + 2 tan θ = 15 θ = 51.34 ◦ , 123.69 ◦ ,
231.34 ◦ or 303.69 ◦
3. 2 cosec 2 t − 5 cosec t = 12
t = 14.48 ◦ , 165.52 ◦ ,
221.81 ◦ or 318.19 ◦
4. 2 cos 2 θ + 9 cosθ − 5 = 0
[θ = 60 ◦ or 300 ◦ ]
15.7 Worked problems (iv) on
trigonometric equations
Problem 14. Solve 5 cos 2 t + 3 sint − 3 =0 for
values of t from 0 ◦ to 360 ◦ .
Since cos 2 t + sin 2 t = 1, cos 2 t = 1 − sin 2 t . Substituting
for cos 2 t in 5 cos 2 t + 3 sin t − 3 = 0 gives:
5(1 − sin
2 t ) + 3 sint − 3 = 0
5 − 5 sin
2 t + 3 sint − 3 = 0
−5 sin
2 t + 3 sint + 2 = 0
5 sin
2 t − 3 sint − 2 = 0
Factorizing gives (5 sin t + 2)(sin t − 1) = 0. Hence
5 sint + 2 = 0, from which, sint =−
2
5 =−0.4000, or
sin t − 1 =0, from which, sin t = 1.
t = sin −1 (−0.4000) = 203.58 ◦ or 336.42 ◦ , since sine
is negative in the third and fourth quadrants, or
t = sin −1 1 = 90 ◦ . Hence t = 90 ◦ , 203.58
◦ or 336.42 ◦
as shown in Fig. 15.7.
1.0
y
t 8
20.4
21.0
0
9 0 8
203.588
y 5 sin t
336.428
2708
3608
Figure 15.7
Problem 15. Solve 18 sec 2 A − 3 tan A = 21 for
values of A between 0 ◦ and 360 ◦ .
1 + tan 2 A = sec 2 A. Substituting for sec 2 A in
18 sec 2 A − 3 tan A = 21 gives
18(1 + tan 2 A) − 3 tan A = 21,