120
Chapter 8 General Solutions of Trigonometric Equations
CHAPTER 8
8. 3 cos θ = 2 sin 2 θ − 3 3 cos θ = 2 − 2 cos 2 θ − 3
Exercise 8A (p.186)
2 cos 2 θ + 3 cos θ + 1 = 0 (2 cos θ + 1)(cos θ + 1) = 0 1 cos θ = − or cos θ = −1 2 2π θ = 2 nπ ± or θ = 2nπ ± π 3 where n is any integer.
1. cos θ = −1 for 0° ≤ θ ≤ 180° θ = 180° π π − 3 for − < θ < 2 2 2 − 3 tan θ = 12 θ = −0.14 (corr. to 2 d.p.)
2. 6 tan θ =
θ = nπ + ( −1) n 0.28 where n is any integer.
θ = nπ −
π 6
1 2
π 4 nπ n π θ= + ( −1) 3 12
3θ = nπ + ( −1) n
where n is any integer. π )= 3 4 π π 2θ − = nπ + 4 3 7π 2θ = nπ + 12 nπ 7π θ= + 2 24
6. tan(2θ −
where n is any integer. 7.
3 + 3 tan 2 θ = 4 tan θ 3 tan 2 θ − 4 tan + 3 = 0 (tan θ − 3 )( 3 tan θ − 1) = 0 1 3 π π or θ = nπ + θ = nπ + 3 6 where n is any integer.
tan θ = 3
where n is any integer. 5. sin 3θ =
3 sec 2 θ = 4 tan θ 3 (1 + tan 2 θ) = 4 tan θ
3. sin θ = 0.28
4. 3 tan θ = − 3 − 3 tan θ = 3
9.
2 cos 2 θ + 3 cos θ + 1 = 0 (2 cos θ + 1)(cos θ + 1) = 0 1 cos θ = − or cos θ = −1 2 2π θ = 2 nπ ± or θ = 2nπ ± π 3 where n is any integer.
or
tan θ =
10. 6 tan θ = 5 csc θ sin θ 5 = 6 cos θ sin θ 6 sin 2 θ = 5 cos θ 6 − 6 cos 2 θ = 5 cos θ 6 cos 2 θ + 5 cos θ − 6 = 0 (3 cos θ − 2)(2 cos θ + 3) = 0 2 3 cos θ = or cos θ = − (rejected) 3 2 θ = 2 nπ ± 0.84 (corr. to 2 d.p.) where n is any integer. 11. sin 5θ = cos 4θ π cos 4θ = cos( − 5θ) 2 π 4θ = 2 nπ ± ( − 5θ) 2 π 4θ = 2 nπ + ( − 5θ) 2 π 9θ = 2 nπ + 2 2 π θ = nπ + where n is any integer. 9 18 or π 4θ = 2 nπ − ( − 5θ) 2 π −θ = 2 nπ − 2 π θ = −2 nπ + 2
Chapter 8 General Solutions of Trigonometric Equations
Since the values of θ =
2 π nπ + include those 9 18
π , the general solution of the equation 2 2 π is θ = nπ + , where n is any integer. 9 18
θ = 2 nπ +
of −2 nπ +
12. tan 4θ = cot 3θ π tan 4θ = tan( − 3θ) 2 π 4θ = nπ + − 3θ 2 π 7θ = nπ + 2 1 π θ = nπ + 7 14
16.
1 (2 cos α ) cos θ − (2 sin α )sin θ = 1 2(cos θ cos α − sin θ sin α ) = 1 1 cos(θ + α ) = 2 π π θ + = 2 nπ ± 3 3 2π θ = 2nπ or θ = 2 nπ − 3
or
θ = 2 nπ ±
π 2
nπ π or θ = 2 nπ ± , where n is any integer. 3 2 tan θ + 3 cot θ = 5 sec θ 5 sin θ 3 cos θ + = cos θ sin θ cos θ sin 2 θ + 3 cos 2 θ = 5 sin θ
sin 2 θ + 3 − 3 sin 2 θ = 5 sin θ 2 sin 2 θ + 5 sin θ − 3 = 0 (2 sin θ − 1)(sin θ + 3) = 0 1 sin θ = or sin θ = −3 (rejected) 2 π θ = nπ + ( −1) n 6
17. sin 2 θ + 1 = 3 cos θ(sin θ + cos θ) 2 sin 2 θ + cos 2 θ = 3 sin θ cos θ + 3 cos 2 θ 2 tan 2 θ + 1 = 3 tan θ + 3
14. Let α be an acute angle such that tan α = 1 . Then 2
5π 12
where n is any integer.
where n is any integer.
2 cos α = 2 2 sin α = 2 π α= 4 The given equation becomes 2(sin θ + cos θ) = 2 sin θ + 2 cos θ = (2 sin α )sin θ + (2 cos α ) cos θ = 2(cos θ cos α + sin θ sin α ) =
15. sin 4θ + sin 2θ = 0 2 sin 3θ cos θ = 0 sin 3θ = 0 or cos θ = 0
θ=
13. Let α be an acute angle such that tan α = 3 . Then 2 cos α = 1 2 sin α = 3 2 π 3 α= 3 α The given equation becomes
or θ = 2 nπ +
where n is any integer.
3θ = nπ
where n is any integer.
π 12
121
2
2 tan 2 θ − 3 tan θ − 2 = 0 (2 tan θ + 1)(tan θ − 2) = 0 1 tan θ = − or tan θ = 2 2 θ = nπ − 0.46 (corr. to 2 d.p.) or θ = nπ + 1.11 (corr. to 2 d.p.) where n is any integer.
α 2
6 3 3 3 3 cos(θ − α ) = 2 π π θ − = 2 nπ ± 4 6
18. cos 2θ = 3 cos θ + 4 2 cos 2 θ − 1 = 3 cos θ + 4 2 cos 2 θ − 3 cos θ − 5 = 0 (2 cos θ − 5)(cos θ + 1) = 0 5 cos θ = (rejected) or cos θ = −1 2 θ = 2nπ ± π where n is any integer.
122
Chapter 8 General Solutions of Trigonometric Equations
19. tan 3θ + sec 3θ = 1 sin 3θ 1 + =1 cos 3θ cos 3θ sin 3θ + 1 = cos 3θ and cos 3θ ≠ 0 cos 3θ − sin 3θ = 1 π 2 cos(3θ + ) = 1 4 1 π cos(3θ + ) = 4 2 π π 3θ + = 2 nπ ± 4 4 π 3θ = 2 nπ or 3θ = 2 nπ − 2 π 2 2 θ = nπ − θ = nπ 3 6 3 The condition cos 3θ ≠ 0 means π 3θ ≠ 2 nπ ± 2 π 2 θ ≠ nπ ± 3 6 2 ∴ θ = nπ , where n is any integer. 3 20. sin 2θ = cos 2θ − sin 2 θ + 1 2 sin θ cos θ = 2 cos 2 θ − sin 2 θ 2 tan θ = 2 − tan 2 θ
When k = 2 n + 1 , π π + 2θ = (2 n + 1)π − − θ 6 4 2n + 1 5π θ= π− 3 36 π 2n + 1 5π π− ∴ θ = 2 nπ + or 12 3 36 where n is any integer.
π π ) + 3 sin(θ + ) = 2 4 4 π π π sin(θ − ) + 3 sin( − + θ) = 2 4 2 4
22. sin(θ −
1 π 3 π sin(θ − ) + cos(θ − ) = 1 2 4 2 4 1 2 3 2 Let r = ( ) + ( ) = 1 , 2 2 3 1 π r cos α = , r sin α = ,∴ α= 2 2 3 π π π π ∴ sin(θ − ) cos + cos(θ − )sin = 1 4 3 4 3 π π sin(θ − + ) = 1 4 3 5π ∴ θ = 2 nπ + where n is any integer. 12 ∴
tan 2 θ + 2 tan θ − 2 = 0 (tan θ + 1)2 = 3 tan θ + 1 = ± 3
tan θ = −1 + 3 or tan θ = −1 − 3 θ = nπ + 0.63 (corr. to 2 d.p.) or θ = nπ − 1.22 (corr. to 2 d.p.) where n is any integer. 21.
2 (cos θ + sin θ) = cos 2θ + 3 sin 2θ π π 2 [ 2 (sin cos θ + cos sin θ)] 4 4 π π = 2 sin cos 2θ + 2 cos sin 2θ)] 6 6 π π ∴ sin( + θ) = sin( + 2θ) 4 6 π π ∴ + 2θ = kπ + ( −1) k ( + θ) 6 4 where k is any integer.
When k = 2 n , π π + 2θ = 2 nπ + + θ 6 4 π θ = 2 nπ + 12 where n is any integer.
23. sin θ + sin 2θ + sin 3θ = 0 (sin 3θ + sin θ) + sin 2θ = 0 2 sin 2θ cos θ + sin 2θ = 0 sin 2θ(2 cos θ + 1) = 0 2 cos θ + 1 = 0 or sin 2θ = 0 1 cos θ = − or 2θ = nπ 2 2π nπ θ = 2 nπ ± or θ = , where n is any integer. 3 2 24. sin 11θ sin 4θ + sin 5θ sin 2θ = 0 1 1 (cos 7θ − cos 15θ) + (cos 3θ − cos 7θ) = 0 2 2 1 (cos 3θ − cos 15θ) = 0 2 sin 9θ sin 6θ = 0
sin 9θ = 0 or sin 6θ = 0 9θ = nπ or 6θ = nπ 1 1 θ = nπ or θ = nπ , where n is any integer. 9 6
Chapter 8 General Solutions of Trigonometric Equations
θ 3θ 25. 4 cos cos =1 2 2 1 4 ⋅ (cos 2θ + cos θ) = 1 2 1 cos 2θ + cos θ = 2 1 2 2 cos θ − 1 + cos θ − = 0 2 4 cos 2 θ + 2 cos θ − 3 = 0 cos θ = 0.651 39 or −1.151 (rejected) ∴
θ = 2 nπ ± 0.86 (corr. to 2 d.p.)
where n is any integer. 26. sin θ sin 7θ = sin 3θ sin 5θ cos 6θ − cos 8θ cos 2θ − cos 8θ = 2 2 ∴ cos 6θ = cos 2θ 6θ = 2 nπ ± 2θ for any integer n. nπ nπ θ= or θ = 2 4 nπ nπ Since the values of θ = include those of , 4 2 nπ the general solution of the equation is θ = , 4 where n is any integer.
29. (a) sin 4 x + cos 4 x = sin 4 x + 2 sin 2 x cos 2 x + cos 4 x − 2 sin 2 x cos 2 x 1 = (sin 2 x + cos 2 x )2 − sin 2 2 x 2 1 2 = 1 − sin 2 x 2 (b) 4(sin 4 x + cos 4 x ) − sin 2 x − 3 = 0 1 4(1 − sin 2 2 x ) − sin 2 x − 3 = 0 (by (a)) 2 2 sin 2 2 x + sin 2 x − 1 = 0 (sin 2 x + 1)(2 sin 2 x − 1) = 0 1 ∴ sin 2 x = −1 or sin 2 x = 2 π n π 2 x = nπ − ( −1) or 2 x = nπ + ( −1) n 2 6 nπ nπ n π n π ∴ x= − ( −1) or + ( −1) 2 4 2 12 where n is any integer. nπ π − ( −1) n , 2 4 let n = 2 m For x =
2 mπ π − ( −1)2 m 2 4 π = mπ − 4 let n = 2 m + 1 x=
π )=3 4 1 1 − tan θ + =3 tan θ 1 + tan θ 1 + tan θ + tan θ(1 − tan θ) = 3 tan θ(1 + tan θ)
27. cot θ + cot(θ +
(2 m + 1)π π − ( −1)2 m +1 2 4 (2 m + 1)π π = + 2 4 (2 m + 1)π π π = − + 2 4 2 π = ( m + 1)π − 4
x=
4 tan 2 θ + tan θ − 1 = 0 tan θ = 0.390 4 or tan θ = −0.640 4 θ = nπ + 0.37
or θ = nπ − 0.57
(corr. to 2 d.p.) (corr to 2 d.p.) where n is any integer.
π 4 π nπ π + ( −1) n ∴ x = nπ − or x = 4 2 12 where n is any integer.
28. sin 3 θ − cos3 θ = sin θ − cos θ
∴
(sin θ − cos θ)(sin θ + sin cos θ + cos θ) = sin θ − cos θ (sin θ − cos θ)(1 + sin θ cos θ) = sin θ − cos θ (sin θ − cos θ)(1 + sin θ cos θ) − (sin θ − cos θ) = 0 (sin θ − cos θ)sin θ cos θ = 0 2
sin θ − cos θ = 0 tan θ = 1 π θ = nπ + 4
2
or or
sin θ cos θ = 0 sin 2θ = 0
or
2θ = nπ n θ= π 2 where n is any integer.
123
30. (a)
x = nπ −
1 [(sin x + cos x )2 − 1] 2 1 = (sin 2 x + 2 sin x cos x + cos 2 x − 1) 2 1 = (2 sin x cos x ) 2 = sin x cos x
124
Chapter 8 General Solutions of Trigonometric Equations
(b) Let t = sin x + cos x , ∴ The given equation becomes 1 t + (t 2 − 1) = 1 2 2 t + 2t − 3 = 0 t = 1 or t = −3 As sin x ≤ 1 and cos x ≤ 1, ∴
t = sin x + cos x = −3 is rejected.
∴
sin x + cos x = 1 π π 1 sin x cos + cos x sin = 4 4 2 π 1 ∴ sin( x + ) = 4 2 π π x + = nπ + ( −1) n 4 4 π n ∴ x = nπ + [( −1) − 1] , where n is any 4 integer.
31. (a) tan θ tan(θ + α ) = k sin θ sin(θ + α ) ⋅ =k cos θ cos(θ + α ) sin θ(sin θ cos α + cos θ sin α ) = k cos θ(cos θ cos α − sin θ sin α ) sin θ cos α + sin θ cos θ sin α 2
= k cos 2 θ cos α − k sin θ cos θ sin α ( k + 1)sin θ cos θ sin α = ( k cos 2 θ − sin 2 θ) cos α k +1 sin 2θ sin α 2 cos 2θ + 1 1 − cos 2θ = [k ( )− ]cos α 2 2 ( k + 1)sin 2θ sin α = ( k cos 2θ + k − 1 + cos 2θ) cos α = [( k + 1) cos 2θ + ( k − 1)]cos α ∴ (1 − k ) cos α = ( k + 1)(cos 2θ cos α − sin 2θ sin α ) = ( k + 1) cos(2θ + α )
π , k = 2 into the result of (a), 3 π π ∴ 3 cos(2θ + ) = − cos 3 3 π 1 cos(2θ + ) = − 3 6 π 2θ + = 2 nπ ± 1.738 3 6n − 1 θ= π ± 0.87 (corr. to 2 d.p.) 6 where n is any integer.
(b) Substitute α =
Revision Exercise 8 (p.188) 1. sec 5θ + csc 2θ = 0 1 1 + =0 cos 5θ sin 2θ ∴ − sin 2θ = cos 5θ ∴
cos 5θ = sin( −2θ) = cos(
π + 2θ) 2
π + 2θ = 2 nπ ± 5θ , where n is any integer. 2 −2 nπ π 2 nπ π θ= + or θ = − 3 6 7 14 2 nπ π 2 nπ π + − ∴ θ= or θ = 3 6 7 14 2. 3 + sin 2 x − sin x = 6 cos x 3 + 2 sin x cos x − sin x = 6 cos x sin x (2 cos x − 1) + 3(1 − 2 cos x ) = 0 (2 cos x − 1)(sin x − 3) = 0 1 cos x = or sin x = 3 (rejected) 2 π x = 2 nπ ± , where n is any integer. 3 π π ) = tan( x + ) 12 12 π) π) sin( x − 12 sin( x + 12 3⋅ = π) π) cos( x − 12 cos( x + 12 π π π π 3 sin( x − ) cos( x + ) = sin( x + ) cos( x − ) 12 12 12 12 3 π 1 π (sin 2 x − sin ) = (sin 2 x + sin ) 2 6 2 6 sin 2 x = 1 π 2 x = 2 nπ + 2 π x = nπ + , where n is any integer. 4
3. 3 tan( x −
1 4. (a) 2 sin x cos( p − x ) = 2 ⋅ [sin p + sin(2 x − p)] 2 = sin(2 x − p) + sin p (b) sin x + 2 sin x cos( p − x ) − sin p = 0 ∴
sin x + sin(2 x − p) + sin p − sin p = 0
∴
sin x + sin(2 x − p) = 0 sin(2 x − p) = sin( − x )
∴ ∴
2 x − p = nπ + ( −1) n ( − x ) [2 + ( −1) n ]x = p + nπ p + nπ x= 2 + ( −1) n
where n is any integer.
Chapter 8 General Solutions of Trigonometric Equations
5. sin x + sin 2 x + sin 3 x + sin 4 x + sin 5 x = 0 sin x + sin 5 x + sin 2 x + sin 4 x + sin 3 x = 0 2 sin 3 x cos 2 x + 2 sin 3 x cos x + sin 3 x = 0 sin 3 x (2 cos 2 x + 2 cos x + 1) = 0 or 4 cos 2 x − 2 + 2 cos x + 1 = 0 or 4 cos 2 x + 2 cos x − 1 = 0
9. (a) Let c = cos 2 x ,
cos x = 0.309 or cos x = −0.809 2π 4π or x = 2 nπ ± x = 2 nπ ± 5 5 where n is any integer. 6. No solution is provided for the H.K.C.E.E. question because of the copyright reasons. 7. (a) sin 2 x + sin 2 2 x + sin 2 3 x 1 − cos 2 x 1 − cos 4 x 1 − cos 6 x = + + 2 2 2 3 1 = − (cos 2 x + cos 4 x + cos 6 x ) 2 2 3 1 = − (2 cos 2 x cos 4 x + cos 4 x ) 2 2 3 1 = − cos 4 x (2 cos 2 x + 1) 2 2 3 (b) sin 2 x + sin 2 2 x + sin 2 3 x = 2 3 sin 2 x + sin 2 2 x + sin 2 3 x − = 0 2 1 − cos 4 x (2 cos 2 x + 1) = 0 2 cos 4 x (2 cos 2 x + 1) = 0 ∴
cos 4 x = 0
or cos 2 x = −
π or 2 nπ π or x= ± 2 8 for any integer n. 4 x = 2 nπ ±
sin θ(cos α − k sin α ) = cos θ( k cos α − sin α )
sin θ(1 − k tan α ) = cos θ( k − tan α ) k − tan α tan θ = 1 − k tan α
sin 2 x =
1 − c 4 1 + c 4 17 2 ) +( ) = c 2 2 16 1 − c 4 1 + c 4 17 2 (b) ( ) +( ) = c 2 2 16 (1 − c) 4 + (1 + c) 4 = 17c 2 (
(c 2 − 2)(2c 2 − 1) = 0 1 ∴ c 2 = 2 or c 2 = 2 2 As c = cos 2 x , ∴ c = 2 is rejected.
2 2 π π or 2 x = (2 n + 1)π ± 2 x = 2 nπ ± 4 4 π 2n + 1 π x = nπ ± ∴ or x = ( 2 )π ± 8 8 where n is any integer. ∴
1 2
2π 3 π x = nπ ± 3
sin θ cos α + cos θ sin α = k (cos θ cos α + sin θ sin α )
1 − cos 2 x 1 − c = 2 2 1 + cos 2 x 1 + c 2 cos x = = 2 2 ∴ (*) becomes
∴
2c 4 − 5c 2 + 2 = 0
2 x = 2 nπ ±
8. (a) sin(θ + α ) = k cos(θ − α )
2 π 3 , α = , ∴ tan α = 3 3 3 2 3− 3 −1 3 1 tan θ = 3 2 = 3 = −1 1− 3 3 ⋅ 3 3 π θ = nπ + , where n is any integer. 6
(b) k =
sin 3 x = 0 or 2 cos 2 x + 2 cos x + 1 = 0 3x = nπ nπ x= 3
125
cos 2 x = ±
10. (a) Let P(n) be the proposition °ßsin θ − sin 3θ + sin 5θ + L + ( −1) n +1 sin(2 n − 1)θ =
( −1) n +1 sin 2 nθ °®. 2 cos θ
When n = 1 , L.H.S. = sin θ ( −1)2 sin 2θ 2 sin θ cos θ = = sin θ 2 cos θ 2 cos θ ∴ P(1) is true. Assume P(k) is true for any positive integer k. R.H.S. =
i.e. sin θ − sin 3θ + sin 5θ + L +( −1) k +1 sin(2 k − 1)θ =
( −1) k +1 sin 2 kθ 2 cos θ
126
Chapter 8 General Solutions of Trigonometric Equations
Then sin θ − sin 3θ + sin 5θ + L + ( −1) k +1sin(2 k − 1)θ + ( −1) k +1+1 sin[2( k + 1) − 1]θ
( −1)
k +1
sin 2 kθ + ( −1) k + 2 sin(2 k + 1)θ 2 cos θ = [( −1) k +1 sin 2 kθ =
(c) sin 2θ =
3 2
π 3 nπ n π θ= + ( −1) 2 6 where n is any integer. 2θ = nπ + ( −1) n
+ 2( −1) k + 2 sin(2 k + 1)θ cos θ] ÷ 2 cos θ 1 = {( −1) k +1 sin 2 kθ + 2( −1) k + 2 [sin 2 kθ 2 + sin 2( k + 1)θ]} ÷ 2 cos θ
12 − 13. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.
= [ −( −1) k sin 2 kθ + ( −1) k sin 2 kθ
Enrichment 8 (p.190)
+ ( −1) k + 2 sin 2( k + 1)θ] ÷ 2 cos θ
( −1)( k +1) +1 sin 2( k + 1)θ = 2 cos θ
Thus assuming P(k) is true for any positive integer k, P( k + 1) is also true. By the principle of mathematical induction, P(n) is true for all positive integers n. (b) sin θ cos θ − sin 3θ cos θ + sin 5θ cos θ + L + sin 9θ cos θ = 0 cos θ(sin θ − sin 3θ + sin 5θ + L + sin 9θ) = 0 cos θ = 0 (rejected) or sin θ − sin 3θ + sin 5θ + L + sin 9θ = 0
( −1)5 +1 sin 2(5)θ =0 2 cos θ sin 10θ = 0 for cos θ ≠ 0 10θ = nπ nπ θ= 10
where n is any integer. 11. (a) Sum of the roots = tan θ + cot θ 4 3 = 3 (b) By (a),
4 3 3 sin θ cos θ 4 3 + = 3 cos θ sin θ sin 2 θ + cos 2 θ 4 3 = 3 sin θ cos θ 3 = 4 3 sin θ cos θ 3 = 2 3 sin 2θ 3 sin 2θ = 2 3 3 = 2 tan θ + cot θ =
1. (a) cos 4θ = 2 cos 2 2θ − 1
= 2(2 cos 2 θ − 1)2 − 1 = 2( 4 cos 4 θ − 4 cos 2 θ + 1) − 1 = 8 cos 4 θ − 8 cos 2 θ + 1 (b) 16 x 4 − 16 x 2 + 1 = 0 1 8x 4 − 8x 2 + = 0 2 Let x = cos θ 1 cos 4θ − 1 + = 0 2 1 cos 4θ = 2
π 3 nπ π θ= ± 2 12 where n is any integer.
4θ = 2 nπ ±
nπ π ± ) 2 12 x = ±0.966, ± 0.259 (corr. to 3 sig.fig.) x = cos(
cos β cos(2α + β) + cos β = cos(2α + β) cos(2α + β) 2 cos(α + β) cos α = cos(2α + β)
2. (a) 1 + m = 1 +
(b) (1 + m) tan α tan(α + β) 2 cos(α + β) cos α sin α sin(α + β) = ⋅ ⋅ cos(2α + β) cos α cos(α + β) 2 sin α sin(α + β) = cos(2α + β) 2( − 12 )[cos(2α + β) − cos β] = cos(2α + β) cos β − cos(2α + β) = cos(2α + β) cos β = −1 cos(2α + β) = m −1
Chapter 8 General Solutions of Trigonometric Equations
(c) Let α = x , β = cos π4
m=
π 4
cos(2 x + π4 ) By (b),
3. tan θ = − 3
θ = tan −1 ( − 3 ) = −60°
=5
k
π ) = 5 −1 4 2 π tan x tan( x + ) = 4 3 1 + tan x 2 tan x ⋅ = 1 − tan x 3 3 tan x + 3 tan 2 x = 2 − 2 tan x
(1 + 5) tan x tan( x +
1
−180° −360° −270°
−90° O
90°
θ 270° 360°
Classwork 2 (p.181)
sin θ =
3 1. sin θ = 2
sin θ +
3 θ = sin ( ) = 60° 2 −1
1 2 3 2
cos θ = −
k
=0
1 2
1 1 tan θ + = 0 3 2
k = sin θ
Principal value
General solution
30°
n 180°n + (−1) 30°
−60°
180°n − (−1)n60°
120°
360°n ± 120°
−56.3°
180°n − 56.3°
θ
O
60° 90°
180°
270°
Classwork 3 (p.183)
360°
1. sin θ + cos θ = 0 sin θ = − cos θ tan θ = −1 π θ = nπ − 4
−1
2. cos θ = −
1 2
where n is any integer.
1 θ = cos −1 ( − ) = 120° 2
2. 1 + 2 3 sin x cos x + 2 cos 2 x = 0
k
sin 2 x + cos 2 x + 2 3 sin x cos x + 2 cos 2 x = 0 sin 2 x + 2 3 sin x cos x + 3 cos 2 x = 0
1 1 2
−1
180°
− 3
Classwork 1 (p.178)
O 1 − 2
−60°
−1
3 tan 2 x + 5 tan x − 2 = 0 (3 tan x − 1)(tan x + 2) = 0 1 tan x = or tan x = −2 3 x = nπ + 0.322 or x = nπ − 1.11 (corr. to 3 sig.fig.) where n is any integer.
1 3 2
127
k = cos θ 120° 90°
θ 180°
270°
360°
(sin x + 3 cos x )2 = 0 sin x + 3 cos x = 0 sin x = − 3 cos x tan x = − 3 π x = nπ − 3 where n is any integer.
128
Chapter 8 General Solutions of Trigonometric Equations
Classwork 4 (p.184)
Classwork 7 (p.186)
tan 4 x − cot 3 x = 0 tan 4 x = cot 3 x π tan 4 x = tan( − 3 x ) 2 π 4 x = nπ + − 3 x 2 nπ π x= + 7 14
1 1 cos 6θ − cos 2θ 2 2 1 1 1 (cos 6θ + cos 2θ) = cos 6θ − cos 2θ 2 2 2 1 1 cos 2θ = − cos 2θ 2 2 cos 2θ = 0 π 2θ = 2 nπ ± , where n is any integer. 2 π θ = nπ ± 4 cos 2θ cos 4θ =
where n is any integer.
Classwork 5 (p.184) π π 1 cos θ − sin sin θ = 5 5 2 π 1 cos( + θ) = 5 2 π π + θ = 2 nπ ± 5 3 π π π π ∴ θ = 2 nπ + − or θ = 2 nπ − − 3 5 3 5 2π 8π or θ = 2 nπ − θ = 2 nπ + 15 15 cos
Classwork 6 (p.185) Let α be an acute angle such that tan α =
1 . Then, 2
1 = 5 sin α 2 = 5 cos α α = 0.463 6 The given equation becomes 5 sin α sin θ − 5 cos α cos θ = 5 cos α cos θ − sin α sin θ = −1 cos(θ + α ) = −1 θ + 0.463 6 = 2 nπ ± π = (2 n + 1)π where n is any integer.
∴
θ + 0.463 6 = 2 nπ + π θ = 2 nπ + 2.68 (corr. to 2 d.p.)