|
- \ | - 1
=
0
3
2
<=>
2
1
^
n k
3 * ~ 2
" T
(2
1
n k \
(
3
ЗтгАгЛ .
в1 1 " Г т п ' =Г т } ‘
e Z .
3.
tg x=a
tenglama
x=
arctg а
+nk, k e Z
yechimga ega bo'ladi. Xususiy
holda
a) agar tg
x —
0
bo'lsa,
x = nk, k e Z
;
n
b) agar tg
x
=
1
bo'lsa,
x = — + nk, keZ;
4
К
d) agar tg
x
=
—1
bo'lsa,
x =—— + nk, keZ;
e) agar tg
2
x= a bo'lsa, x=±arctg^/a
+nk, keZ.
Misol. 3
tg2 3x
- 1 = 0
tenglamani yeching.
Y e c h i s h .
\t g23x =
1
3x
=
±arctg
+
kn
V3
<=> ^3x = ± ^ +
kn
j
4. ctg
x
=
a
tenglama
x
= arcctg
a + nk, ke Z
yechimga ega bo'ladi.
К
a) agar ctgx =
0
bo‘lsa,
x = — як, ke Z
;
к
b) agar ctgx = 1 bo‘lsa, x = — +
nk
,
keZ;
/1
d) agar ctgx
= —1
bo'lsa, x = - — + тг&,
keZ;
e) agar ctg2x =
a
bo'lsa, x =±arcctg>/a +
як, keZ.
Misol.
ct£ 2
^2x - —
j
= 3 tenglamani yeching.
211
<=>
(
, л
л
л к \
.
о х = ± — + —+ — ,
к е z-
12
6
2
/
Matematika kursida har qanday trigonometrik tenglamalar ayniy almash-
tirishlarni bajarish orqali taqqoslanib sinx#a, cosx=a,
tgx=a,
ctg
x=a
ko'ri-
nishdagi eng sodda trigonometrik tenglamalarga keltiriladi.
Trigonometrik tenglamalar quyidagi metodlar yordamida yechiladi.
1
. Ко‘ paytuvchilarga keltirish usuli.
1-misol.
sin
2
x = cos
2
x sin
2
x tenglamani yeching.
Y e c h is h . sin2x - cos2x sin2x = 0, sin2x(l-cosx) =0
1) agar l~cosx*0 bo'lib, sin2x=0 bo'lsa,*\= у
n, n e Z
bo'ladi.
2) agar sin2x*0 bo'lib, 1—cosx=0 bo'lsa, cosx=l, x=2
nn, n e Z
bo'ladi.
2-misol.
sin3x — sin x = 0 tenglamani yeching.
Y e c h i s h . sin3x - sin x = 2sin x cos 2x = 0
1) agar cos2x*0 bo'lib, sinx=0 bo'lsa,
x=nn, ne.Z\
7t
И7С
2)
agar sinx
*0
bo'lib, cos
2
x
= 0
bo'lsa,
x = ~
4
+ ~ ^ ’ n e Z
bo'ladi.
3-misol.
cos
2
x+cos
2
2x+cos
2
3x= 1,5 tenglamani yeching.
,
1
+ c o s
2
x
„
Y e c h is h . cos2x = ----- ------ formulaga ко ra
l
+ cos2x
l + cos4x
l
+ c o s
6
x
3
«>
<=> (co s
2
x + cos4x + c o s
6
x =
0
)
[(c o s
2
x + c o s
6
x) + cos4x =
0
] <=>
<=> [2 c o s4 x c o s2 x + cos4x =
0
] <=> co s4 x (c o s2 x +
1
) =
0
.
%
m
1) agar 2cos2x+l*0 bo'lib, cos4x=0 bo'lsa, x= — + — , n e z ;
o
4
2) agar cos4x*0 bo'lib, 2cos2x+l=0 bo'lsa, c o s 2 x = -^ -,
2л ^
к
„
2х
=— — + 2пп
; x = - - + jr
П ; neZ.
II.
0
‘zgaruvchilarni kiritish usuli.
1-misol.
2cos
2
x=3 sin x tenglamani yeching.
Y e c h is h . (2cos2x — 3 sin x = 0 )» (3 sin
x —
2(l-sin
2
x) = 0
о (3sin
x — 2 +
2 sin2x =0).
Agar sin
x = у
desak,
2y
2
+ 3y - 2 = 0,
У г
= -2 .
s in x = i j<=> | x =
7
~ + 2kn
j ,
k e z .
2-misol.
cos2x — 5 sin
x —
3 = 0 tenglamani yeching.
Y e c h i s h . cos2x= l-2sin2x formulaga ko‘ra
(1
-2sin
2
x-5sin x -3 = 0 ) о
(2sin
2
x+5sinx+2=0) « sinx=,y desak,
2.У
2
+
+ 2 = 0, j , = —2,
1
* — 2-
1
) sin
x ~ —2
tenglama yechimga ega emas.
\
, k e z;
2
) ( s in x = - - | o x = ( - l ) fc a r c s in f - -
1
+яА:
I
2J . I ■
I 2J J
x = (-1
)*+1
^ +
nk, k e Z.
6
III. Bir jinsli tenglamalarni yechish.
1- misol.
2
sin
2
x-sinxcosx—cos
2
x
= 0
tenglamani yeching.
Y e c h i s h . Bu tenglama sinus va kosinus funksiyalariga nisbatan bir
jinslidir. Tenglamalarning har ikki tom onini cos
2
x*0 ga bo'lsak,
2tg2x — tg x - 1 = 0 hosil bo‘ladi. Bundan tgx= 1 va tg x = — ^ .
л
1
) agar tgx=l bo'lsa,
x = - — +nk, k e Z
;
2) agar tg x = — ^ bo'lsa, x = —arctg^-
+nk, ke Z
bo'ladi.
2- misol.
cos
2
x+3sin
2
x+2 ^3 sinxcosx=3 tenglamani yeching.
Y e c h is h . Bu tenglama ayniy almashtirishlar bajarish orqali bir jinsli
ko'rinishga keltiriladi.
213
cos
2
x +
3 sin
2
x +
2л/3 sin
x
cos
x =
3(sin
2
x +
cos
2
x),
cos
2
x + 3sin
2
x + 2>/3 s in x cos x - 3 sin
2
x - 3 cos
2
x = 0
2
cos
2
x - 2%/3 sin x cos x =
0
.
2
cos x(cos x - л/3 sin x) =
0
.
1
) agar cosx - V3sinx/0 bo‘lib, cosx
=0
bo'lsa,
х = ^ + л к , keZ;
2
) agar cos x
*0
bo'lib, cosx - л/3 sin л=Ь bo'lsa, t g x = ^ , x = ^ r +&
7
Г,
keZ;
IV. asin x+bcosx =
с
ko'rinishdagi tenglamani yechiftg.
X
/ hsu/. Bu tenglamani yechish uchun /&— = / almashtirish bajariladi.
2
f c §
Ma’lum ki,sin * ~ ~
^ x ’
c o s x ~
"
2
* ’ edi, shunga ko'ra berilgan
l + fc
2
1
+ #
2
tenglama quyidagi ko'rinishni oladi:
2
аГ + 6 0 - ^ ) =Ci
2
at + b - b t 2 =c + ct2,
\ + t2
1
+
12
7
a ± yla2 + b2
—
c2
(b + c)t - 2at + (c - b)
=
0
,
t
= ------------- ---------
c +
b
~
d + v a
2
+
b2 — с2
л
,
~>
i
'
i
i
i
x =
2
arctg
---------------------- +
2
Att,
k e z , a + b > c
va
b ± - c .
c + b
Agar
6
= -c bo'lsa, kvadrat tenglama chiziqli tenglamaga almashadi:
A
b
2at+2b=0,
t
= ---- , x = -2arctg — +2for, &e Z
о
a
I I usul.
Tenglamaning har ikkala tom oniл/а
2
+ й
2
ga bo'linadi:
2
<
1
.
a
b
rSinx + -F
. . - - cosx
yja2
+
b2
л/я
2
+
b2
j a 2 + b2
214
Va
2
+
b2
= 1 va
* 1 ,
a
b
Agar
r n
— .T - cos ^ va
Va
2
+
b2
\la 2
+
b
2
sin ^ desak, berilgan tenglama
с
sinx • cos
r ~
r ko'rinishni oladi, bundan sin (x+
vfl +
0
С
\y
X Va2 +
b2
ko‘ladi. p =arctg
— ;
agar
a ^ b 2^. c1
bo'lsa, x
= (- ! ) *
arcsin
7 + jzk-a rctg — ,
k e z -
Va
2
+
ь2
' a
1-misol. 3cosx + 4sin x = 5 tenglamani yeching.
Y e c h i s h .
^32
+
42
_
^25
b o 'lg an i uch u n tenglam aning h a r ikki
3
4
Ш
2
f 4 f
to m o n i 5 ga b o 'lin a d i: - c o s x + — s in x = 1, I -
+ l ' j I =
s h u n in g
3
4
uchun - = sin
va — = cos
bo'ladi, bundan sin
cosx+cos
tenglama hosil qilinadi yoki sin(x+
1
bo'ladi:
x + (p = ?L + 2 kn,
x = ^ +
2
kit - (p,
k e z ,
. 3
л
. 3
,,,
.
arcsin - ,
x = - - a r c s i n - +
2кк,
k e z -
-X
X
2
X
2
tg -
1
-tg -
2- usul.
Agar sin
x
= ------ -— , va cos x ---------- -, ekanligini nazarda
2
x
l + «
2
x
1
-
У
2
2
v
tutib,
tg -г = У
desak, 3 • ------
j +
4
' 1-----
2
= ^
,yoki
3~3y2+8y-5+5y2
yoki
2
1
+ v
1
+
у
215
4y
2
-4y+ l= 0 bundan
у = ^
yechim hosil bo‘Iadi:
f x
1 \
f x
1
^
1
( / ^ 2
=
2
arctg2 + л к \' x ~ larctg —
+
2nk,
k e n .
MUSTAQIL YECHISH UCHUN MISOLLAR
Trigonometrik tenglamaiarni yeching.
3!
I. tgx +
sin
xtgx
= 0.
Javobi: nn.
2 -2 s i n x - 3 co s
x
=
6
.
Javobi
: x e 0 .
3. sin
2
x - (
1
+ >/3,) s in x co sx + л/3 cos
2
x =
0
.
К
Javobi: — + nn,
arctgl + пл.
•V
7t
4-
sin
2
x - 4 sin x c o s x + 3 cos
2
x = 0.
Javobi:—
+
nn,
arctg3 + nn.
5- л/зsin
2
x - 4 s in x c o s x + V3 cos
2
x = 0.
Javobi: — + nn,
^ + k n .
j
о
5
J3
it
6
. sin2 x + 3cos
2
x - 2 s i n x c o s x = — - — .
Javobi: y + kn.
2
6
7
_
t
_
j
t
—
"s/53
'■
7 cos x - 7 sin 2x = 2.
Javobi: x = arctg
----- ------.
2
8
. т /X .
7 = 1-
Javobi: x =
(-1)* •
~ + kn, k e Z-
3 v 2 s i n x - l
4
9-
tgx - 2 = ^ ~
^ x'
Javobi: kn, ^~ + кл.
10
. (
1 - 2
sin x) sin x =
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