158 Higher Engineering Mathematics
i.e. 18 + 18 tan
2 A − 3 tan A − 21 = 0
18 tan
2 A − 3 tan A − 3 = 0
Factorizing gives (6 tan A − 3)(3 tan A + 1) = 0.
Hence 6 tan A − 3 =0, from which, tan A =
3
6 = 0.5000
or 3 tan A + 1 =0, from which, tan A = −
1
3 = − 0.3333.
Thus A = tan −1 (0.5000) = 26.57 ◦ or 206.57 ◦ , since
tangent is positive in the first and third quadrants, or
A = tan −1 (−0.3333) =161.57 ◦ or 341.57 ◦ , since tangent is negative in the second and fourth quadrants.
Hence,
A = 26.57
◦
, 161.57
◦
, 206.57
◦ or 341.57
◦
Problem 16. Solve 3 cosec 2 θ − 5 = 4 cot θ in the
range 0 <θ < 360
◦ .
cot 2 θ + 1 = cosec 2 θ. Substituting for cosec 2 θ in
3 cosec 2 θ − 5 =4 cot θ gives:
3 (cot
2
θ + 1) − 5 = 4 cot θ
3 cot
2
θ + 3 − 5 = 4 cot θ
3 cot
2
θ − 4 cot θ − 2 = 0
Since the left-hand side does not factorize the quadratic
formula is used. Thus,
cot θ =
−(−4) ±
[(−4) 2 − 4(3)(−2)]
2(3)
=
4 ±
√
(16 + 24)
6
=
4 ±
√
40
6
=
10.3246
6
or −
2.3246
6
Hence cot θ = 1.7208 or −0.3874, θ = cot −1
1.7208 =30.17 ◦ or 210.17 ◦ , since cotangent
is positive in the first and third quadrants, or
θ = cot −1 (−0.3874) = 111.18 ◦ or 291.18 ◦ , since
cotangent is negative in the second and fourth quadrants.
Hence,
θ = 30.17
◦
, 111.18
◦
, 210.17
◦ or 291.18
◦
Now try the following exercise
Exercise 69 Further problems on
trigonometric equations
In Problems 1 to 12 solve the equations for angles
between 0 ◦ and 360 ◦ .
1. 2 cos 2 θ + sin θ = 1
[θ = 90 ◦ , 210 ◦ , 330 ◦ ]
2. 4 cos 2 t + 5 sin t = 3
[t = 190.1 ◦ , 349.9 ◦ ]
3. 2 cosθ − 4 sin
2
θ = 0
[θ = 38.67 ◦ , 321.33 ◦ ]
4. 3 cosθ + 2 sin 2 θ = 3
[θ = 0 ◦ , 60 ◦ , 300 ◦ , 360 ◦ ]
5. 12 sin 2 θ − 6 = cos θ θ = 48.19 ◦ , 138.59 ◦ ,
221.41 ◦ or 311.81 ◦
6. 16 sec x − 2 = 14 tan 2 x
[x = 52.53 ◦ or 307.07 ◦ ]
7. 4 cot 2 A − 6 cosec A + 6 = 0
[A = 90 ◦ ]
8. 5 sec t + 2 tan 2 t = 3
[t = 107.83 ◦ or 252.17 ◦ ]
9. 2.9 cos 2 a − 7 sina + 1 =0
[a = 27.83 ◦ or 152.17 ◦ ]
10. 3 cosec 2 β = 8 −7 cot β
β = 60.17 ◦ , 161.02 ◦ ,
240.17
◦ or 341.02
◦
11. cot θ = sin θ
[51.83 ◦ , 308.17 ◦ ]
12. tan θ + 3 cot θ = 5 secθ
[30 ◦ , 150 ◦ ]
i.e. 18 + 18 tan
2 A − 3 tan A − 21 = 0
18 tan
2 A − 3 tan A − 3 = 0
Factorizing gives (6 tan A − 3)(3 tan A + 1) = 0.
Hence 6 tan A − 3 =0, from which, tan A =
3
6 = 0.5000
or 3 tan A + 1 =0, from which, tan A = −
1
3 = − 0.3333.
Thus A = tan −1 (0.5000) = 26.57 ◦ or 206.57 ◦ , since
tangent is positive in the first and third quadrants, or
A = tan −1 (−0.3333) =161.57 ◦ or 341.57 ◦ , since tangent is negative in the second and fourth quadrants.
Hence,
A = 26.57
◦
, 161.57
◦
, 206.57
◦ or 341.57
◦
Problem 16. Solve 3 cosec 2 θ − 5 = 4 cot θ in the
range 0 <θ < 360
◦ .
cot 2 θ + 1 = cosec 2 θ. Substituting for cosec 2 θ in
3 cosec 2 θ − 5 =4 cot θ gives:
3 (cot
2
θ + 1) − 5 = 4 cot θ
3 cot
2
θ + 3 − 5 = 4 cot θ
3 cot
2
θ − 4 cot θ − 2 = 0
Since the left-hand side does not factorize the quadratic
formula is used. Thus,
cot θ =
−(−4) ±
[(−4) 2 − 4(3)(−2)]
2(3)
=
4 ±
√
(16 + 24)
6
=
4 ±
√
40
6
=
10.3246
6
or −
2.3246
6
Hence cot θ = 1.7208 or −0.3874, θ = cot −1
1.7208 =30.17 ◦ or 210.17 ◦ , since cotangent
is positive in the first and third quadrants, or
θ = cot −1 (−0.3874) = 111.18 ◦ or 291.18 ◦ , since
cotangent is negative in the second and fourth quadrants.
Hence,
θ = 30.17
◦
, 111.18
◦
, 210.17
◦ or 291.18
◦
Now try the following exercise
Exercise 69 Further problems on
trigonometric equations
In Problems 1 to 12 solve the equations for angles
between 0 ◦ and 360 ◦ .
1. 2 cos 2 θ + sin θ = 1
[θ = 90 ◦ , 210 ◦ , 330 ◦ ]
2. 4 cos 2 t + 5 sin t = 3
[t = 190.1 ◦ , 349.9 ◦ ]
3. 2 cosθ − 4 sin
2
θ = 0
[θ = 38.67 ◦ , 321.33 ◦ ]
4. 3 cosθ + 2 sin 2 θ = 3
[θ = 0 ◦ , 60 ◦ , 300 ◦ , 360 ◦ ]
5. 12 sin 2 θ − 6 = cos θ θ = 48.19 ◦ , 138.59 ◦ ,
221.41 ◦ or 311.81 ◦
6. 16 sec x − 2 = 14 tan 2 x
[x = 52.53 ◦ or 307.07 ◦ ]
7. 4 cot 2 A − 6 cosec A + 6 = 0
[A = 90 ◦ ]
8. 5 sec t + 2 tan 2 t = 3
[t = 107.83 ◦ or 252.17 ◦ ]
9. 2.9 cos 2 a − 7 sina + 1 =0
[a = 27.83 ◦ or 152.17 ◦ ]
10. 3 cosec 2 β = 8 −7 cot β
β = 60.17 ◦ , 161.02 ◦ ,
240.17
◦ or 341.02
◦
11. cot θ = sin θ
[51.83 ◦ , 308.17 ◦ ]
12. tan θ + 3 cot θ = 5 secθ
[30 ◦ , 150 ◦ ]