460
CHAPTER 8 Analytic Trigonometry
Double-Angle, Half-Angle, and
Product-Sum Formulas
1. sin x = !J, x in quadrant]
cos x = H and tan x = -fu. Thus, sin 2x = 2 sin x cosz = 2 ( fl) (H) = i~g,
sin 2x
i~g
cos2x =cos
sin x = (13) - (13) =~ =
and tan2x =--- = 119 =
ill =
cos 2x
2. tan J; =- ~. Then sin x = t and cos x =- ~ (x is in quadrant II). Thus, sin 2x = 2 sin x cos x = 2 . t (- ~ )
sin 2x
- ~
cos 2x =cos x - sin x = (-"5) - ("5) = ~ =- 2"5' and tan 2x = --2- =
= 2"5 . """7 = """7'
cos x
-2"5
3. COS,I; = ~. Then sinx = -~ (cscx < 0) and tanx = -~. Thus, sin2x = 2sinxcosx = 2 (-t) . t = -~,
sin2x
-~
- and tan2x =- - =-- =--.
-~7'!'
cos 2x
15
'7
4. cscx = 4. Then sinx = ~, cosx = - vps, and tanx = -:.;h (tanx < 0). Thus,
~, and
sin Zz = 2sinxcosx = 2· ~ (- VPS) =- v'J5", cos2x = cos x - sin x =(- ~f - G/
sin
Vis
tan 2x =_,"-- =__ =_v'15 . =_v'15.
cos 2x
~
5. sinx = -to Then, cos x = -~ and tan x = ~ (x is in quadrant III). Thus, sin 2x =2 sinx cos x =2 (-t) (-~)
. 2
cos 2x =cos
sirr' x =(_ *)
(_.:?) = 1~- = .J..., and tan 2x = cos
sm x = 25 = ~ . .:!..5. ~-= ~.
2x
15
6. sec z = 2. Then cosx = ~, sinz = - ~, and tanx = -J3 (x is in quadrant IV). Thus,
sin' x
( ~)
(~~ f = - ~, and
sin 2x = 2 sin x cos x = 2 ( - ~) (~) = - V;, cos 2x = cos
8.3
=}
.
2
.
2
12
2
5
2
144-25
119
169'
J; -
120
169
169'
11
12
120
119'
169
2
.
2
3
2
4
2
9-16
7
24
25
24
13
-7-
7
24
25'
25
2
')x
25
14
2
-
§
8_
8
7
7
24
2 X -
2 __
2
"
~5
5
9
25
25
2
7
1')
2 -
X -
sin2x
- ~
=- = --=J3.
tan 2x
7. tanx
7
-~
cos 2x
>
-~ and cosx
=
< O.
0, so sinx
k. Thus,
(kf-(-Jof=fo
1
=
-:.;To and cos x
Thus, sinx
=2 sin x cos x =2 ( - Jo) (.;h) =- fa =- ~, cos 2x =cos sin x =
sin 2x
- ~
and tan 2x =--- =- =- - .- =- -.
cos 2x
.'!
8. cotx = ~. Then tanx = ~,sinx = .;T3 (sin x > 0), and cos x = k. Thus,
. 2 x =2'
SIn
sin x cos x = 2 ( .;T3
.;T3 = 13' cos 2x = cos x - SIn x = (kf-(kf= ~~ = -fl, and
, tan 2x = sm -x = _13 = 11 . (_11) =__ 11
cos2x
-!3
2
sin 2x
5
3
5
3
5
4
4
2
X -
1!
21
2'
3
3
'')
)
(
•
'j
.
2
9 sin
x -- (sm 2 x) -
-
=4: 1
4
10. cos x
1
~ (~
=
(cos x)
_ 14
-
(1 --
cos
1
'2 cos 2x -\- 4: .
=
12
2
.
2
2
5
2X)
5'
2
2
--41
-
1 -\- cos 4x
2--
.
,
- 12 cos 2x -\- 14
cos 2 2x
= 4: 1
1
.
1
1
'2 cos 2x -\- '8 -\- '8 cos 4x
= '8 3
1
_
1
'2 cos 2x -\- '8 cos 4x
- cos2x -\- ~ COS4X)
2
2
=
(1 -\-
2X)
cos
--2--
1 cos 2x -\- 14 .
-\- 2
2
12
13
.c
2)
2
1
.1:....± cos
2 4x __
- 14
= ~ (~ -\- cos 2x -\- ~ cos 4x)
1
1
2
4: -\- 2' cos2x -\- 4: cos 2x
1 cos 2x' -\- 18
-\- 2
1
cos 4x
- 28
-\- 8
1 cos _x
'). -\- 1
8 cos 4x
-\- 2
461
SECTION 83 Double-Angie, Half-Angle, and Product-Sum Formulas
11. We use the result of Example 4 to get
2
COS
X
.
4
, 2 ,
( sin x
sm x
cos 2 x )
, 2 X
Sill
16 (1 ~ cos 2x -
2)
= (
81
- 81
cos4x) .(21
- 1
2 cos
x
-+ cos 2x cos 4x)
cos 4x
12. Using Example 4, we have
4
,2
cos xsm x
cos '
J'
2
(cos .r sin"
16 (1 13. Since sin 4xcos4 x = (sin 2 z cos '
,4
X)2
cos 4x
x) = ~ (1 -l- cos2x)· ~ (1- cos4x)
-+ cos 2x
- cos 2x cos 4x)
we can use the result of Example 4 to get
1 81 cos4x )2 = 64
1 - 32
1 cos4x
( 8-
4
SIn xcos x
~ - ~ cos 4x -l- ~ , ~ (1
1~8
.-
~ cos 4x
-+
1 cos 2
4x
-+ 64
-+ cos 8x)
Th- cos 8'£ =
= ~ - ~ cos 4x -l- 1~8 -l- 1~8 cos 8:£
~ (~ - cos 4x
-+
~ cos 8x)
14. Using the result of Exercise 10, we have
(i -l- ~ cos 2x -l- ~ cos 4x) 0 -l- ~ cos 2x;)
-& -+ ~ cos 2x -+ 16 cos 4x -+ -& cos 2x -l- 16 cos 2x cos 4x
-& -+ ~ cos 2.'£ -+ 16 cos 4x -+ 16 cos 2x cos 4;y, = 16 (3 -+ 7 cos 2x -+ cos 4x -+ cos 2x cos 4x)
2
(cos x) 2 . cos
=
16. tan 15°
= -- cos_ =
1
30°
=
X
F0 -1)
J~ (1- cos300) =
15. sin 15°
2
1 _ V3
- - 1 -2-
sin 30°
= 2-
2
= Jf(2 - V3)
~J2 - V3
=
V3
_~cos45° = ~ - -~ = J2-1
17. tan 22.5° =
sin4,'jo
V2
2
18. sin 75°
J~ (1~;I5c)o) =
=
19. cos 165°
=
-J~ (1 -+ cos 330°) =
20. cos112.5° =
21. tan
i
j
-/f(i -+ cos22S
1 - cos
7r.
0
)
1 _ v'2
= -~-'!. = __2_ =
7r.
8
sin ~
37T
I·
{:(1- [-
22. cos '8
-/f(1
=
1]) = Jf(2 -+ V3) = ~J2 -+ V3
-+ cos30~
=
-j ~
(1
--/~ (l-=-cos45°) = -j~
-+
~)
(1-
=
-~J2 -+ V3
"i) = -~J2= J2
J2 - 1
V2
2
11
1.37T,
-+ cos :3f
= cos (2 . "4) = V- - 2 - -
(1 -
vi'
~ J2= J2. Note that we have chosen the
= \ -2­
a; is in quadrant T, so cos ~7T > O.
23. cos:& = ~ (1 -+ cos i) = ~ (1 ; 1) = ~ J 2 -+ V3
positive root because
24. tan
_
'0 =
1 - cos
..
25. sin ~
=-
sin
.J7T
,-6-
0;
=
1 -l-
v::
--1-"-
"2
= 2 -l- V3
/f(i .- cos 9;) = -
¥ < O.
26. sin \1~~ = vi ~ (1 - cos 1~
quadrant II, so sin -¥:f > O.
j~
(1 - ,;)
=-
~ J2 -
J2. We have chosen the negative root because
9 71'
8
is in
quadrant III, so sin
'!. )
~
J2 -- J3. We have chosen the positive root because
\\; is in
462
CHAPTER 8 Analytic Trigonometry
(b) 2 sin 3& cos 3&
2 tan 7°
28. (a) - - - - 1 - tan? 7°
= sin 36°
27. (a) 2 sin 18° cos 18°
= sin 6&
sin8°
1 + cos 8°
1 - cos 4&
(b) -----sin4&
8 0
2
4&
tan 2
(b) 2sin
31. (a) - - - - = tan-- = tan4°
=
32 . (a
) \
= tan 2&
+ x) = sin x cos x + cos x sin x = 2 sin x cos x
35. sin x
=
Since x is in quadrant I, cos x
.
x _
sm"2
-- V/12" (1 - cos x ) -_
x
tan "2
36. cos z
90
.
0
2
==
37. esc x
0
sin ~
cos "2
=
--x
-2--
34. tan (x
+ x) =
. x_
cos "2 -
.
2
=
"S 90° and so
~
~
cos "2x -_
"S 45° and so
cos -a2
# "S
=-
=-
_1_
Vli5
= sin4&
tan x + tan ~
1 -- tan x tan x
+ cos z_) -_ V/12"
V!l(1
2"
(1
2V2) _
----:;- -
and cosr =
~
J
21
= ~
(
cos x ) --
=
2
= -
J~ (1 -
2~.
=
2tanx
1 - tan? x
+ 54) =
_3_
Vli5
(1
=
j1;. (1 -
:7.:
+ 54) --
"S 270°,
3 -Vli5
~~ = ~
. Vli5 =
Vli5 -1
J~
(1
+
r.»
_ a: _ sin ~ _
2y 2 , and tan "2 - cos ~ -
~
"S
h::J]
10'
-3.
2[2)
cos x) =
=
V~ (3 + 2\1'2),
3+2V2
~-2J2 = 3
F(
"S x "S 90°,
V;)
1-
+ 2yr.>2.
=
~ -J2-- \1'2,
V2
r.> d.
2 - -_ y r.>
1 + 2V2) --- 21 V~2
L, + V 2, an
tan "2"' _- 1 - 'cos
. x_I
- - -- =
2 - 1.
sm z
= -
and since x is in quadrant TV, sin x
# is in quadrant
II. Thus,
V::.
v2
2
Since 270
0
sin~ = J~ (1- ~osx) = V~
"S x "S 360
(1-
0
we have
,
n = Jti =
2
2
v%6
-~
3
Jz
~
2
V;:,
<
v5
esc x
> 0 and
< 0, it follows that :r is in quadrant Ill. Thus 180° "S x 'S 270° and so 90 "S ~ "S 135
Thus
~
Then, cosx =
-
and sinx =
-
(cscx
is in quadrant U. sin
~
=
V~ (1 -
cosx)
.= - V (1 + cos x) = - ~(----)
/
=
V26
1.
2
.1
2
1-..J5..~
_1.
2
=
J~
(1
26-5V26
13'
+
-~)
and tan
.5 .
Since cotx
0).
0
2
10'
(f0(
sin~ = ---.",
1 ..-v'6 = - -ro
1 = - V5
. 1 + cos x ) = - V
-1 (1 + -2) = - -V5
= - V30,.
and tan - = ­
V6
6'
cos
vs -V5
40. cotx = 5.
cos f
~Vli5 and
=
Since 90° "S :r "S 180°, we have
=
cos:r)
sin
cos
Vh:2
v;, since :r is in quadrant T. Also, since 0°
is also in quadrant 1. Thus, sin
180° and so
10 '
=
/1 (,.3 V"6
~. Also, since 180°
Vli5 and tan I.
and since x is in quadrant II, cos x
V;
=
~
5
is in quadrant 1. Thus, sin
/f(:(
2" 1 + cos x )-
,39. sec x = ~. Then cos x
135° "S
~
/1 (1 + cos x_) -_ V:2
!1 ( 1 V:2
38. tan x = 1. Then sin x
0° "S
0
-3­
3. Then, sin x
~
'S
.
= sin If
~ and ~ is also in quadrant 1. Thus,
=
(1 - 54) _
1 -_ 1iJ,
y'lO cos "2
x ~
- Vli5
.-
V (1 + cos--;;:) = - V!l (1- :!) = 2
o
= sin
v'l6
1
Vli5 .
1.
~ cos ~
/1 -- cos 30°
= -~. Since x is in quadrant Ill, sinx = -~ and tanx =
<
_·.r -. /1
-:2x -< 1<]50
.) and so "2x IS· ·III quadrant II ..Th us, SII1"2
V
2" (1 -
COS'"
45
=
V/12"
. 2 &
sm 2 = cos&
J1-- ~OS8&
(b)
33. sin (x
~.
2& -
2
30. (a) cos
(b) cos 2 5& - sin' .5& = cos 10&
0
2 tan ~ = tan 14&
1 - tan? 7&
(b)
29. (a) cos 2 34° - sin" 34° = cos 68°
= tan 14
I.
2
=
0
•
~ ~+t,v2G,
l-cosx
=--sin x
~~ 1
+ -~
Jzv",6 =
-
-5 -
V26.
463
SECTION 8.3 Double-Angle, Half-Angle, and Product-Sum Formulas
41. sin2x cos 3:r = ~ [sin (2x -I- 3x) -I- sin (2x - 3x)] = ~ (sin Sz - sin x)
42. sin x sin 5x
=~
[cos (x - 5x) - cos (x -I- 5x)]
=
~ (cos 4x - cos 6x)
43. cos x sin 4x = ~ [sin (4x -I- x) -I- sin (4x - x)] = ~ (sin 5x -I- sin 3x)
= ~ [cos (5x -I- 3x) -I- cos (5x - 3x)] = ~ (cos 8x -I- cos 2x)
45. 3 cos 4x cos 7x = 3 . ~ [cos (4x -I- 7 x) -I- cos (4x - 7 x)] = ~ (cos 11x -I- cos 3x)
46• 11 sin '2£ cos '4£ c= 11 . 102 [sin ('£2 -I- '£)
-I- sin ('£2 - '4,
£ 'I] = D2 (sin 3x4 -I- sin '£)
4
4
44. cos 5x cos 3x
. 3 x = 2'
5x -I- 3X) cos (5X
- 3X) = 2'
. 5 x -I- sin
47· sm
sm ( -~2---2-~
SIn 4 x cos x
48. sin x - sin 4x
=
2 cos ( x
~ 4~ )
sin (
:l::-~ 4X) = 2 cos
· (4X -I. (4X
- 6X)
2 sin
2 6X) sin
--2--
49· cos 4x - cos 6 x
=-
50. cos 9x -I- cos 2x
= 2 cos ( 9X -I-2 2X)
' 2x
51· sm
. (2X
- 7 x) =
="
cos( 2x -I2 7 x) sin
--2--
~
. 7x
sm
cos
¥
sin ( _ 3
=-
2X)
= _ 2 cos
2 sin.
. 5XSIn
. ( -x )
=
5; sin 3;
2'
.
.sin 5 xsmx
(9X ~
2X) = 2 cos 2
11x cos 2
7x
2
L,
9x sin
. (
2 cos 2
~
25X) = - 2 cos 29x sm. 25x
.7x
.7x
. 4x = 2 sin
. (3 X-l-4X)
52· sin 3x -I- sin
2
cos (3X-4X)
2
= 2 sin
2 cos (X)
- '2 = 2 sin
2 cos '2x
53. 2 sin 52.5° sin 97.5° = 2 . ~ [cos (52.5° - 97.5°) - cos (52.5" -I- 97.5°)] = cos (-4.5°) - cos 150°
= cos 45°
54. 3cos37.5° cos 7.5°
55. cos 37.5° sin 7.5"
=~
=~
-- cos 150°
= v'2
-I2
=
102
(h -I- vS)
(v:; -I- 1) = ~ (h -I- vS)
sin Su") = ~ (v:; -~) = ~ (h -1)
(cos4.5° -I-cos300)
(sin45° -
V3
2
=~
56• sin 75° -I- c
sin 15°
- 2 sin (75° -I. V3 - y'6
' 2 15°) cos (750 -2 150) -- 2 sin 45° cos 30° -- 2. v'2
22-2
. (2550 -I. (2550
-.19.5°) = - 2'
v'2) 21
57· cos 255 ° - cos 19.5° = -2 sm
2 195°) sin
--~2--sm2 250'
sin 300 = -2' ( - 2
,T -I- cos 12
571' = 2 cos [12 (71'
)] cos [12 (12
71' - 12
5")] =- 2 cos _71'4 CDc~ _71'6 -_ 2. v'2
. V3
-_
58· cos 12
12 -I- 571'
12
2
2
59. cos" 5x - sin' 5x = cos (2. 5x) = cos lOx
60. sin 8x
= sin
61. (sinx -I-
(2 . 4x)
COSX)2
=
2 sin4x cos 4x
= sin' x -I- 2sinxcosx -I- cos" x = 1+ 2sinxcosx = 1 -I- sin2x
2 tan x
2 tan x
sin x
2
•
62.
2
= - - 2 - = 2 . - - cos x = 2 sin x cos x = sm 2x
1 -I- tan x
sec x
cos x
63. sini:z: = 2sin2xcos2x = 2 (2sinx~osx) (cos2x) = 4cosxcos2x
sin x
sin x
sin x
1 -I- sin 2x __ 1 -I- 2 sin x cos x
1 -I1
_
64.
= 1 -I- ~ csc x sec x
sin 2x
2 sin x cos x
2 sin x cos x
2
65 2(tanx-cotx)
2(tanx-cotx)
· tan 2 x - c o t2x
(tanx-l-cotx) (tanx-cotx)
tan x -I- cot x
sin x cos x
2
. ---sin x
cos x sin x cos x
---I--~­
cos x
SIn x
1 - tan 2 x
1
1
- - = ---;~-­
--­
2tanx
tan2x
2tanx
1 - tan 2 x
-,------
66. cot 2x
=
2sin x cos x
. 2
2
sin x -I- cos x
=
2
sinx
cos x
---1--.­
cos x
sin :r
.
.
2 sin x cos x = SIn 2x
y'6
2
~=
2v'2
464
CHAPTER 8 Analytic Trigonometry
tan2x + tan x
67. tan Sz = tan (2x + x) =
=
1 - tan2xtanx
2tanx
1 ---t-a-n-'2'-x- + tan x
2tanx
') tanx
1 - tan- x
1-
3tanx - tan 3 x
1 - 3tan 2 x
2tanx + tanx (1 - tan 2 x)
1 - tan? x - 2tan:rtanx
68. 4 (sin6 x + cos 6 x) = 4 [(sin 2 x + cos" x)3 - 3 (sin 4 xcos 2 x + sin 2 z cos" x)]
= 4 [1- 3 sin" x cos 2 x (sin 2 x + cos ' x)] = 4 - 12 sin" x cos" x
= 4 - 3 (2 sinx cos X)2 = 4 - 3sin2 2x
69. cos" x - sin" x = (cos 2
o
+ sirr' x) (cos 2
X
7r
2 +"4
70. Let y
tan"
X
(X-
2
7r)
+ 4
{=}
sin 2 x) = cos"
X -
X
12
1 - cos 2y
= tan y =
=
' 1 + cos 2y
1+
Then
+ 2'
cos (x + ~)
11+
cos (x +~)
sin x + sin 5:1:
2 sin 3x cos 2x
sin 3x
= - - - - - - = - - = tan3x
cos x + cos 5x
2 cos 3x cos 2x
cos 3x
72.
sin 3x + sin 7tc
cos 3x - cos 7x
73.
sin lOx
sin 9x + sinx
74.
sin x + sin 3x + sin 5x
cos x + cos 3:y; + cos 5x
sin z + siny
cos x + cosy
sin 2 x = cos 2x
7r
2y
71.
75.
X -
(- sinx)
(- sinx)
1 + sinx
1- sinx
2 sin .5x cos 2x
cos 2x
2
= - - = cot' x
-2 sin 5x sin (-2:1:)
sin 2x
---------c-----,-
2 sin 5x cos 5x
2 sin 5x cos 4x
cos5x
cos4x
sinx+sin5x+sin3x _ 2sin3xcos2x+sin3x _ sin3x(2cos2x+1)
---:----',-,------e- = tan 3x
cos x + cos 5x + cos 3x
2 cos 3x cos 2x + cos 3x
cos 3x (2 cos 2x + 1)
y) (:;:-y)
x+
2 cos ( -2cos (X-V)
-2-'
sin(~)
x2
+
- cos - 2 2 sin ( y )
cos
y)
,(x+y+x2 sm
2
sin (x + y) - sin (x - y)
76. cos (:r + y) + cos (x - y)
C;Y)
cos
X +y)
tan ( 2­
(x+Y-x+ y)
2
2 cos ( x + y + x -- y ) cos (x+Y-:;;+Y)
2
2
'
o
.
° = 2 cos--130° + 110° , 130
77 . sm130
-sm110
2--sm
0
•
°
2 110° = 2 cos 12 00'
sm1 00 =2 ('-2I)'
sm10 0 =-sm10
-
0
0
100° + 200° SIn
, -----=--100 - 200 = -'2 Bl.n
' 150° Sl.Il. ( - 5 0 0)
Sl Il
78 . cos 100 e, - cos 20 0° = - 2'
2
= - 2 (~) (- sin 50 0
79. sin 45° + sin 15° = 2 sin (
2
)
= sin 50°
(450 -- 15°)
45° + 1.5°)
2
cos
2
= 2 sin 30° cos 15° = 2 ' ~ . cos 15°
= cos 15° = sin (90° - 1.5°) = sin 75° (applying the cofunction identity)
80. cos 87° + cos 33° = 2 cos
87° + 33°
87° - 33°
2
cos
2
= 2 cos 60° cos 27°
= 2· ~ cos 27° = cos 2T = sin (90° - 27°) = sin 63°
sin x + sin 2:x; + sin 3x + sin 4x + sin 5x
81. cos x + cos 2x + cos 3x + cos 4x + cos 5x
(sin x + sin 5x) + (sin 2x + sin 4x) + sin 3x
(cos x + cos 5x) + (cos 2x + cos 4x) + cos 3x
2 sin 3x cos 2x + 2 sin 3x cos x + sin 3x
2 cos 3x cos 2x + 2 cos 3x cos x + cos 3x
sin 3x (2 cos 2x + 2 cos x + 1)
= tan3x
cos 3x (2 cos 2x + 2 cos x + 1)
473
SECTION 8.5 Trigonometric Equations
57. (a)
o = sin- 1
o = sin-I
58.
(2·3
1
+ 1) tan 10°
= 2, () = sin-I (
(b) Forn
and
(
(
1
)
9 tan 15°
~
1
3 tan 15°
f (x) = sin
(sin- 1
= sin- 1
)
1
)
5 tan 15°
~
1__ )
7tan 100
( __
48.3°. Forn
~ sin- 1 0.8102 ~ 54.1°
= 3, o = sin- 1
(
1
)
7 tan 15°
~
= 4,
32.2°. Forn
~ 24.5°. n = 0 and n = 1 are outside of the domain for (3 = 15°, because ~~1_ ~ 3.732
tan 15°
1.244, neither of which is in the domain of sin -I.
x) = x has domain [-1, 1] and
9 (x)
= sin-I (sin x) has domain ( -00, (0) and range
[-~' ~]. Note that 9 (x) = x only for x E [-~' ~].
range [-1,1].
y
y
x
-27J
<tr
7T
59. (a) Let () = see 1 x. Then sec () = x, as shown in the figure. Then cos () =
() = cos -1
(b) Let I
-1
(~). Thus csc-
1
(~), x
x = cos -I
= esc -1 x. Then esc I =
I = sin-I
(e) Let (3
(~). Thus sec
x
2' 1.
x, as shown in the figure. Then sin I
=
(~), x
sin-I
.!.,
so
x
=
2' 1.
= cot- 1 x. Then cot (3 = z, as shown in the figure. Then tan (3 =
(3 = tan":'
(~). Thus cot-
1
x = tan-
1
.!.,
so
x
(~). x
.!.,
so
x
2' 1.
x
Trigonometric Equations
1. cos x + 1 = 0 <=?
for any integer k.
cos x
=
-1. In the interval [0, 21f) the only solution is x
+ 1 = 0 <=? sin x = ~ 1.
x = 3; + 2kIT for any integer k.
2. sin x
=0
solutions are x =
3. 2 sin x-I
4./2 cos x-I
2 sin x
<=?
~
=0
solutions are x = ~
=1
=
In the interval [0, 21f) the only solution is x
<=?
sin x
=
=
IT. Thus the solutions are x
=
3;.
(2k
+ 1) IT
Therefore, the solutions are
~. In the interval [0, 21f) the solutions are x
=
~'
56" .
Therefore, the
+ 2k1f, 56" + 2kIT, k = 0, ±1, ±2, ....
<=?
V2 cos x = 1
<=?
cos x
=
+ 2kIT, 7: + 2kIT for any integer k.
~. The solutions in the interval [0, 21f) are x
=
~'
7:. Thus the
474
CHAPTER 8 Analytic Trigonometry
5. -J3 tan x + 1 = 0
-J3 tan x = ~ 1
9
Therefore, the solutions are x =
6. cot x + 1 = 0
9
v;
tan x = -
In the interval [0, 7f) the only solution is x =
56".
+ km , k = 0, ±1, ±2, ....
56"
cot x = -1. The solution in the interval (0, 7f) is x = 34". Thus, the solutions are x = 3; + k7f for
9
any integer k.
7.4cos 2x-1=0
4c:os2x=]
9
i,
the solutions are x =
23"
,
4;,
53".
4c:os2x=1
9
c:os 2x=%
9
So the solutions are x =
i
c:osx=±~.IntheintervaI[0,27f)
9
+ 2k7f, 23" + 'Ikrn. 43" + 2k7f,
i + k n, 2; + kt: for any integer k,
2
c:os X = ~ 9
c:os x = ±
9
x = ~, 3;,
5;
+ 2k7f, which can be
expressed more simply as x =
B. 2 cos" x- 1 = 0
J:z
9
5;,
7; in [0, 27f). Thus, the solutions are
%+
x =
kn, 34" + kat for any integer k,
9. sec' x - 2 = 0 9
sec 2 x = 2 9
sec: x = ±V2. In the interval [0, 27f) the solutions are x = ~, 3;,
the solutions are x = (2k + 1) %for any integer k.
10. c:sc: 2 X - 4 = 0 9
c:sc: 2 X = 4 9
esc x = ±2. In the interval [0, 27f) the solutions are x =
solutions are x = ~ + kn,
11. 3 esc 2 x - 4 = 0
5; +
56'"
7;, 1~ or. So the
ktt for any integer k.
c:sc:2
x =4~
9
i,
5;, 7;. Thus,
2
esc: x = ± V3
9
sin x = ± Y2'3
9
X
{-c}
4"
= ~, 2
3'" 3'" "; for x in
[0, 27f). Thus, the solutions are x = i + ktt , 2; + k» for any integer k,
12. 1 - tan 2 x = 0 9
tan 2
x = 1 9
tanx = ±1 9
x = ~, 3; in (0, 7f). Thus, the solutions are ~ + km ,
3; +
k.t: for any integer k,
13. c:osx(2sinx+1) =0
3; and sinx =
-~
cosx=Oor2sinx+1=0
9
sinx=-~.On[0,27f),cosx=0
9
x = 7;, 1~". Thus the solutions are x = ~ + km, X =
9
7
6"
9
x=~,
+ 2k7f, X = 1~" + Zk:»; k = 0,
±1, ±2, ....
14. sec x (2 cos z - V2) = 0
2cosx - V2 = 0
for any integer k.
sec x = 0 or 2 cos x - V2 = 0. Since [sec z 2': 1, sec x = 0 has no solution. Thus
9
2cosx = V2
{-c}
15. (tan x + -J3) (cos x + 2) = 0
-i +
V;
x = ~, 7; in [0, 27f). Thus x = ~ + 'Ikrn, 7; + 2k7f
{-c}
tan x + -J3 = 0 or cos x + 2 = 0. Since [cos xl -s: 1 for all x, there is no solution
9
for cos z + 2 = 0. Hence, tan x + -J3 = 0
.r =
cos x =
9
9
tan x = - -J3
9
:£
= -
i
on (- ~, ~). Thus the solutions are
ktt , k = 0, ±1, ±2, ....
16. (2cosx+-J3)(2sinx-1)=0
9
2cosx+-J3=00r2sinx--1=0
9
cosx=-V;orsinx=~
9
x = ~, 5;, 7; in [0, 27f). Thus, x = ~ + km, 56" + 2k7f for any integer k,
17.cosxsinx-2cosx=0
cosx(sinx-2)=0
9
cosx=00rsinx-2=0.Sincelsinxl-S:1forallx,
there is no solution for sin x - 2 = 0. Hence, cos x = 0
±1, ±2, ....
9
x = ~ + 2k7f,
. 1B. tan x sin x + sin x = 0
9
sin x (tan x + 1) = 0
9
{c~
3;
+ 2k7f
9
x = ~ + lctt, k = 0,
sin x = 0 or tan x + 1 = 0. Now sin x = 0 when x = k7f
3
3;
and tan x + 1 = 0 9
tan x = -1 =? x = 4" + k n . Thus, the solutions are x = krn,
+ kt: for any integer k.
2x-4cosx+1=0
19.4cos
9
(2cosx-1)2=0 9
2cosx-1=0 9
cosx=~
9
x=i+2k7f,
5; + Zl:«, k =
0, ±1, ±2, ....
20. 2sin 2 x - sinx - 1 = 0
2sinx+1=0
9
9
(2sinx + 1) (sinx -1) = 0
2sinx=-~
9
{-c}
2sinx + 1 = 0 or sinx - 1 = 0. Since
x=7;,1~"in[0,27f)andsinx=1
9
x=~in[0,27f).Thusthe
solutions are x = '; + 2k7f, l{"- + 2k7f, ~ + 2k7f for any integer k.
21.sin 2x=2sinx+3 9
sin 2x-2sinx-3=0 9
(sinx-3)(sinx+1)=0 9
sinx-3=00r
sin x + 1 = 0. Since [sin x I -s: 1 for all x, there is no solution for sin x - 3 = 0. Hence sin x + 1 = 0 9
sin x = -1
9
x =
3; + 2k7f, k: =
0, ±1, ±2, ....
SECTION 8.5 Trigonometric Equations
22.3tan 3,r:=tanx
3tan 3 x - t a n x = 0
<=?
Now tan x = 0
x = kit and 3tan 2 x-I = 0
==?
~'C + ktc. Thus the solutions are x = k-n, ~ + k n ,
2
23. sin ,r = 4 - 2 cos
2
tanx(3tan 2 x - l ) =0
<=?
2
sin x + cos
<=?
X
for all .r, it follows that cos" x
24.2cos 2x+sinx = 1
2
56"
+ cos
X
2
t.arr' x = ~
<=?
tan x = ± ~
x = ~ + kst ,
==?
+ ktt for any integer k,
= 4
X
1 + cos 2
<=?
:s; 1 and so there is no solution for cos
2
2(I-sin 2 x ) + s i n x - l = 0
<=?
tanx = 0 or3tan 2 x-I = O.
<=?
==?
475
= 4
X
cos 2
<=?
X
= 3 Since leas xl
:s;
1
x = 3.
-2sin 2 x + s i n x + l = 0
<=?
<=?
2 sin 2 x - sin x-I = 0 which we solved in Exercise 20. Thus x = 7; + 2kn, 1~" + 2kn, ~ + 2kn for any integer k,
25. 2 sin 3x + 1 = 0
2~", 3~'C, 3~"
2 sin 3x = -1
<=?
x = ~~,
<=?
26. 2 cos 2x + 1 = 0
27. sec 4x - 2 = 0
\IS", 1;;, 2{S", 215;, 2;;.
cos 2x = - ~
<=?
sec4x = 2
<=?
sin 3x = - ~. In the interval [0, 6n) the solution are 3x = 7;, 1~", 1~",
<=?
2x =
<=?
i
4x =
<=?
2
3"
So X = ~~ + 2k %'
W+ 2k %for any integer k.
+ Zkrn ,
<=?
+ Zl:n ,
4
3"
53"-
+ 2kn
+ 2kn
i
x =
+ ktt ,
+ ktt for any integer k.
2;
x = f§- + ~ km , - f§- + ~ k-t: for any
<=?
integer l: .
. 28. V3tan3x + 1 = 0
29.
J3 sin 2x
+ ki
~
x =
~
32. 2 sin
<=?
tan 2x = ~ (if cos 2x
5;
3x =
<=?
oF 0)
+ kr:
x = i~ + ~ kn for any integer k.
<=?
2x = ~ + ktt
<=?
x = f§- + ~ k.r: for any integer k.
<=?
_._1_ = sin3x <=? sin 23x = 1 <=? sin 3x = ±1 <=? 3x = ~ + kt:
<=?
sm3x
for any integer k. Notice that multiplying by sin3x in the first step did not introduce extraneous roots.
<=?
- 1= 0
O£
3
tan 3x = -- ~
= cos 2x
30. esc 3x = sin :3x
31. cos
<=?
+ V3 = 0
~
cos
<=?
= 1
2 sin
<=?
'--3"
~
<=?
= 2kn
= - V3
x = 4kn for any integer k.
<=?
sin "'3'
<=?
-
3
v'3
2
<=?
O£
3
=
+ 2kn ' 5"
+ 2kn
3
4"
3
x = 4n + 6kn,
<=?
5n + 6kn for any integer k.
33. tan ~ + V3 = 0
34. sec ~ = cos ~
<=?
5
35. tan ,r: - 9tanx =
tan x = ±V3
36.3tan
tan ~ = -
<=?
cos ~ = 1
2
°
==?
37. 4 sin x cos x + 2 sin x - 2 COS;:I: -1 = 0
sin x = ~ or cos x = -~
<=}
4x-9)
<=?
+ kJr <=?
s; + 4kn for any integer k.
x =
x = 2kn for any integer k.
<=?
= 0 or tan 4 x = 9
= 0
in [0, 2n). Thus, x =
(tan.T-l)(3tan
<=?
x = ~, ~,
<=?
23"
cos ~ = ±1
i, 2;, 4;, 5;
3J;-3tan2 x - t a n x + l = 0
tan x = 1 or tan x = ± ~
~ =
<=?
tanx(tan
<=?
x = 0, n,
<=?
J3
k
for any
3"
2x-l)=0
integer k.
(2 sin x-I) (2 cos x + 1) = 0
<=?
x = ~,
tanx=lor3tan 2 x = 1
<=?
for x E [0, n). Thus x = ~ + k-n, ~ + kit ,
56'"
5;, 2
3",
tanx = 0 or
<=?
<=?
+ kr: for any integer k.
56"
2 sin x-I = 0 or 2 cos x + 1 = 0
<=?
~'C in [0, 2n). Thus x = ~ + 2kn,
56"
+ Zkrn ,
2
3"
+ Zktt,
.'if + 2kn for any integer k.
2 sin 2x
sin 2x = - - cos2x
cos2x = 2 (which is impossible since Icosul
38. sin 2J: = 2 tan 2x
<=?
39. cos 2x - sin 2x ~~ 0
2
2
sin 2x cos 2x - 2 sin 2x = 0
<=?
:s;
1) or sin 2x = 0
<=?
2x = kn
(cos 2x - sin 2x) (cos 2x + sin 2x) = 0
<=?
sin 2x (cos 2x - 2) = 0
<=?
x = ~kn for any integer k.
<=?
cos 2x = ± sin 2x
<=?
<=?
tan 2x = ±1
<=?
2x = ~, 3:, 5:, 7:, 94", 1~", 1~", 1~" in [0,4n)
<=?
x = if, 3s", 5;, 7;, 9S", 1~", 1;". 1;"- in [0, 2n). So the
solution can be expressed as the odd multiples of if which are x = if + k4" for any integer k.
<=}
40.secx-tanx = cosx
2
1- sin:r; = 1 - sin x
or sin x = 1
<=?
cosx(secx-tanx) = cosX(COS.T)
<=?
<=?
2
sinx = sin x
<=?
I-sinx = cos 2x
<=?
2
sin x - sinx = 0
<=?
sinx (sin x - 1) = 0
<=?
<=?
sinx = 0
x = 0, n or x = ~ in [0, 2n). However, since the equation is undefined when x = ~, the solutions
are x = kTr for any integer k.
41. 2cos3J: = 1
<=?
cos 3x = ~
<=?
csc
==?
3x =
i,
53'" 7;, 1~", 1~", 1~" on [0, 6n)
<=?
x =
ii,
5;, 79" , 1~", 1~"-,
l~" on [0, 2n).
2
42. 3csc x = 4
2
X
= ~
<=?
cscx = ±
3:J
<=?
sinx = ±
l'
<=?
x =
%' 2;, 4;, 5;
in [0, 2n).
476
CHAPTER 8 Analytic Trigonometry
43. 2 sin x tan x - tan x = 1 - 2 sin x
2 sin x tan x - tan x + 2 sin x-I = 0
B
2 sin x-I = 0 or tan x + 1 = 0
B
sin x = ~ or t.an z = -1
B
i,
x =
B
(2 sinz - 1) (tan x + 1) = 0
B
567f or x = 3;, 7;. Thus, the
i, 3;, 5;, 7;.
solutions in [0, 2n-) are
1 sin x
cos x
sin x
cos ' x
cos x . -.sinx B
- - - -cos x cosx
smx
cos? x
sin x
Multiplying both sides by the common denominator cos" ;r sin x gives sin 2 x - cos" X = sin 2 x cos" x
44. secxtanx - cosxcotx = sinx
B
sinx.
---- -
sin' x - cos" X = (1- cos x) cos 2 X
2
sin:' x - cos" X = cos 2 X - cos" X
B
B
sin 2 x = cos 2
X
B
B
sin' x = 1 - sin' x B
2 sin'' x = 1 B
sin x = ± ~
B
x = %' 3;, .5;, 7;. Since we multiplied the above
2
equation by cos x sin x (which could be zero) we must check to see if we have introduced extraneous solutions. However,
%,
each of the values of x satisfies the original equation and so the solutions on [0, 2n-) are x =
45. tan x - 3 cot x = 0
1 - 4cos 2 X
= 0 B
cos x sin x
x = i, 237f , 437f , 5; in [0, 2n-).
1 - 4 cos''
B
46.2sin 2x-cosx=1
2
sin x - 3cos
cos X sin x
sinx _ 3cosx = 0
sin .z
cos X
B
B
X
= 0
X
B
5 7f , 7 7f
4
4
.
2
B
0
cos x
B
-2cos 2x-cosx+1=0
2cosx-1=00rcosx+1=0
B
,
sin ' x + cos" X - 4cos X
------c----- =
cos X sin x
= 0
4 cos 2 X = 1
B
2(1-cos2x)-cosx-1=0
(2cosx-1)(cosx+1) =0
2
3 7f
4
cosx=~orcosx
B
-1
B
i,5;, n- in [0, 2n-).
x =
47. tan3x + 1 = sec3x
(tan3x + 1)2 = sec 23x
=?
2
2
sec 3x + 2 tan 3x = sec 3x
tt , 4
5;
3",
2 tan 3x = 0
B
B
3;r = 0, n-, 2n-, 3n-, 4n-, 5n- in [0, 6n-)
B
B
x = 0,
i,
2;,
i, n-, 5; are not solutions.
Hence the only solutions are
2;, 4 .
3"
48.3sec 2x+8cos 2 x = 7
4 cos"
B
B
in [0, 2n-) .Since squaring both sides is an operation that can introduce extraneous solutions, we must check each
of the possible solution in the original equation. We see that x =
x = 0,
tan 23x + 2tan3x + 1 = sec 23x
X -
7 cos"
X
3
--2-+4cos2x-7=0(forcosxfO)
cos x
+ 3= 0 B
(4 cos 2 X - 3) (cos 2 x-I) = 0 B
B
B
x =
i,¥,
B
3+4cos 4 x - 7 c o s 2 x = 0
4 cos 2 X
-
3 = 0 or cos 2 x-I = 0
6 , 1~7f or x = 0, n- in [0,2n-). So the solutions are
7"
49. (a) Since the period of cosine is 2n- the solutions are x ~ 1.15928 + 2kn- and x ~ 5.12391 + 2kn- for any integer k.
(b) cos x = 0.4
=?
x = cos- 1 0.4 ~ 1.15928. The other solution is 2n- - cos- 1 0.4 ~ 5.12391.
50. (a) Since the period is n-, the solutions are of the form x
(b) 2 tan x = 13
x
~
B
1.41815 + n-
~
tan x = l~
4.55974.
=?
B
cos x =
i
=?
1.41815 + kr: for any integer k.
x ~ 1.41815. Since the period for tangent is n-, the other solution in [0, 2n-] is
51. (a) Since the period of secant is 2n-the solutions are x
(b) sec x = 5
~
x = cos- 1
~
1.36943 ± 2kn- and
~
4.91375 ± 2kn- for any integer k.
i ~ 1.36944. The other solution is 2n- -
52. (a) Since the period is n-, the solutions are of the form x
~
i ~ 4.91375.
1.16590 + lett for any integer k.
(b) Since cos x = 0 is not a solution, we divide both sides by cos x. 3 slnz = 7 cos x
X ~
cos- 1
B
tan x =
i
B
1.16590. Since the period for tangent is n-, the other solution in [0, 2n-] is x ~ 1.16590 + n- ~ 4.30750.
477
SECTION 8.5 Trigonometric Equations
~
53. (a) Since sin (Jr + x) = - sin x, we can express the general solutions as x
any integer k.
(b)5sin 2x-1=0
sin 2x=i
=?
=?
sinx=±)s.Nowsinx=)s
x
~ Jr - 0.46365 ~ 2.67795. Also, sinz = -)s
x
~
Jr - (-0.46365)
~
3.60524 and x
~
¢}
cos x (4 sin x-I) = 0
¢}
x
~
x = sin-
=?
2Jr - 0.46365
54. (a) Since the period is 2Jr, the solutions are of the form x
any integer k.
(b) 2sin2x - cos x = 0
~
~
cos .c = 0 or 4 sin
1: -
1
( -
=?
)s)
~
2.67795 + k-n for
x=sin-l()s)~0.46365,and
~ -0.46365, so in the interval [0, 2Jr),
.5.81954.
1.57080 + kn ;»
2(2sinxcosx) - cos x = 0
¢}
0.46365 + kn and x
¢}
~
0.25268 + 2kJr, x
~
2.88891 + 2kJr for
4sinxcosx - cosx ~~ 0
1 = O. Now cos x = 0
¢}
¢}
x = ~, ~7f and 4 sin x-I = 0
sin x = ~ =? x ~ 0.25268 or x ~ Jr - 0.2.5268 ~ 2.88891. Thus the solutions in [0, 2Jr) are
0.25268, 1.57080,2.88891,4.71239.
55. (a) Since the period for sine is 2Jr, the general solution is of the form x ~ 0..33984 + 2kJr and x ~ 2.80176 + 2kJr for any
integer k.
(b)3sin 2x-7sinx+2=0
=? (3sinx-1)(sinx-2)=0 =? 3sinx-1=Oorsinx-2=0.Since
[sin z ::; 1, sinx - 2 = 0 has no solution. Thus 3sinx - 1 = 0
x
~
Jr - 0.33984
~
=?
sinr = ~
=?
x ~ 0.33984 and
2.80176.
56. (a) Since the period is Jr, the solutions are of the form x ~ 1.10715 + k n , x
J: ~ 1.89255 + k.r: for any integer k.
(b) tan 4x-1.3tan 2x+36 =0 ¢}
(tan2:r-4)(tan2x~9)=0
~
2.03444 + ktt ,
¢}
J; ~
1.24905 + kn ,
tanx=±20rtanx=±.3
=?
x ~ 1.10715, -1.10715, 1.24905, -1.24905 in (- ~, ~). Since the period for tangent is Jr, the solutions in [0, 2Jr)
are x
~
1.10715, 1.24905, 1.89255, 2.03444,4.24874,4.39064,5.03414,5.17604.
57. f(x)=3cosx+1;g(c)=cosx-1.f(x)=g(x)when
3 cos x + 1 = cos x - I
¢}
2 cos x = - 2
¢}
cos x = - 1 ¢}
x ~~ tt + 2kJr = (2k + 1) Jr. The points of intersection are ((2k + 1) zr, -2) for
any integer k.
58. f(x) = sin2x; g (x) = 2sin2x + 1. f (x) = g (x) when sin2x = 2sin2x + 1
{=?
sin 2x = -1
¢}
21: =
So the intersection points are
59.
3; +
(34
7f
2kJr
¢}
X
=
3; + k.tt for any integer k.
+ k n , -1), for any integer k.
f (x) = tan x; g Cr) = v'3. f (x) =
g
(x) when tanx =
v'3 ¢}
x = ~ + kit .The intersection points are (~ + ktt , v'3) for any integer k.
y
478
60.
CHAPTER 8 Analytic Trigonometry
f (:1:) = sin z
f (x) = 9 (x) when sinx - 1 = cosx =;.
2
sin x - 2sinx + 1 = cos? X
<=?
-1; 9 (x) = cosx.
(sinx - 1)2 = cos 2 X
<=?
sin' x - 2 sin x + 1 - cos 2
X = 0 <=? 2 sirr' x - 2 sin x = 0
2 sin x (sin x-I) = 0
<=?
%+
sin x = 0 or sin x = 1 <=? x = kn,
<=?
2kIr.
However, x = kt: is not a solution when k is even. (The extraneous solutions were
introduced by squaring both sides.) So the solutions are x = (2k + 1) zr,
%+
Zk:«,
and the intersection points are (7f + 2k7f, -1), (~ + 2k7f, 0) for any integer k,
Another method: sin x -1 = cos tc
V; cos x
sin (x
sin ,r - cos x = 1
<=?
~ cos x) = 1
-
<=?
V; sin x­
= V;
<=?
V;. Since sin ~ = cos ~ = V;, we may write V; sin x - V; cos x = cos ~ sin x - sin ~ cos x
-~) = V; <=? tc - if = if, 3: <=? x = %, 7f in [0, 27f). So the solutions are x = (2k + 1) zr, %+
=
and the intersection points are (7f + 2k7f, -1),
61.cosxcos3x-sinxsin3x=0
1;",
V2 ( ~ sin x
<=?
1~" in [0, 87f)
'D
,
cos4x=0
<=?
<=?
63. sin 2x cos tc + cos 2x sin x = 2v!3
<=?
x=~, 2;, 7g
T , 8;,
64.sin3xcosx-cos3xsinx=0
,.
cos (x - 2x) = "2
<=?
cos (-x) = "21
v!3
sin (2x + x) = --:2
<=?
v!3
sin3x = --:2
1
1;", l~" in [0, 27f).
sin(3x-x)=0
<=?
x =, 0, ~, zr, 32" in [0, 27f).
65.sin2x+cosx=0 <=? 2sinxcosx+cosx=0
x ==!!- 7n- 31T 1111"
4x=%,32",5;,72",9;,1~",
<=?
in [0 27f)
3" 5" 7" 9" 11" 13" 15"
8'8'8'8'8'8'8'8
62. cos a; cos 2x + sin x sin 2x = "2
in [0, 27f).
<=?
+ Zkm , 0) for any integer k,
=?!:.
1
in [0,67f)
(%
cos(x+3x) = 0
<=?
Zk:«,
sin2x=0
<=?
cosx = "21
<=?
5"
x= 3'
" :3
2" , 7" 8" 13" , 14"
3x = 3'
" :3
:3 :3 :3 :3
<=?
<=?
<=?
2x=0,7f,27f,37f,47fin[0,47f)
<=?
2' 6'
2'
,r.
0
66. tan -2 - sin x =
sin x
<=?
1 + cosx
<=?
.
sin x = 0
-
<=?
sin x - sin x (1 + cos x) = 0 (and cos x
<=?
i=
-1
x
i= 7f)
x = 0, ~, 32" in [0, 27f) (x = 7f is inadmissible).
2
<=? 2 cos X + cos x - 3 = 0
<=?
sin x = 0 or cos x = 0 <=?
2 cos 2 x - I + cos x - 2 = 0
<=?
67. cos 2:1: + cos x = 2 <=?
(2cosx + 3) (cosx -1) = 0
cosx=l
cosx=Oorsinx=-~
<=?
6
sin x (- cos x) = 0
<=?
cos,r(2sinx+l)=0
<=?
2cosx + 3 = 0 or cos x -1 = 0
<=?
cos x = -~ (which is impossible) or
x=Oin[0,27f).
sinx
cos x = 8 sin
. x cos x
(sinx
cosx) . sin
. x cos x = (
- + -.-<=?
- - + -.8.
sm x cos x ) .
cosx smx
cos x
sin z
2(2sinxcosx)2
<=?
(sin2x)2 = ~ <=?
sin 2x+cos2x = 8sin2xcos 2 x <=? 1
68. tan z-j-cot z = 4sin2x
sinxcosx
<=?
sin 2x = ± ~. Therefore. 2x =
[0, 27f) the solutions are x =
9;, 1~",
1~",
1;",
-if,
if +
3: +
k7f or 2x =
5;, 9;,
1~" and
which are odd multiples of
i.
3;, 7;,
ktt
<=?
1~",
1;".
3;
-if
+ k2" or a; =
+ h~". Thus on the interval
Together we write the solutions as x = -if, 38", 5;, 78",
x =
So we can express the general solution as x =
-if
+ kif where k is any
integer.
69. sin x + sin 3x = 0
2x
=
kr: or
X
=
<=?
k% <=?
70. cos 5x - cos 7 x = 0
61'; = kt: or x = k7r
%or 3,r = i
x =
5\~
+ %k'ir.
2 sin 2x cos x = 0
sin2x = 0 or cosx = 0
x = k% for any integer k.
-2sin6xsin(-x) = 0
<=?
sin 6.1' sin x = 0
sin6x = 0 or sinx = 0
x = k% for any integer k.
71.cos4x+cos2x=cosx
<=?
2 sin 2x cos ( -x) = 0
<=?
2cos3xcosx=cosx
+ 2k7f, 5; + 2k7r,
7
3"
cosx(2cos3x-l)=0
<=?
+ 2k7f, 1~" + 2k7r, 1~" + 2k7f, 1 ~" + 2k7r
<=?
<=?
cos x = 0 or cos 3x = ~
x= ~
+ k7r, ~ + %kJr,