|
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 =
2
cos 2x -
1
.
Javobi:
x = 2
k n ~ .
2
sin
4
x + cos
4
x -
2
s in
2
x + sin
2
2
x =
0
.
Javobi: x
=
arcsin (2 -
J 2
) + ^ .
216
12
. sin
3
x cos x - c o s
3
s in x = cos4 — .
Javobi: x '■
13. sin
4
x
+ cos
4
x =
cos
4x.
14.
tg(
40° +
x)ctg(
5° - x ) = - .
Javobi: x ~
1
(~
1)*+1
arcsin - +
kn
Javobi: — n.
n k
- 35° + (-l)* +l arcsin —
v
’
10
15.(sinx+cosx) =tgx+ctgx.
16. sin3x-sin
3
x+cos3xxos
3
x ;
17. sinx+sm3x+sin7x=3,
18.
tg7x+tg3x-0,
19. 7+sinx+cosx=Q,
29
20
. sm,
0
x+cosl
0
x
es7 7
cos
42
x.
lo
8
'
К
Javobi: x*=2kn+— .
4
Javobi: x= n k ± ^
•
о
Javobi: x=0.
Javobi: x ■
n k
10
'
n
Javobi: х}= кк~— , x2=n(2n+l)
T
A -
2 k + l
Javobi: x -
— -— rt.
21
. Itgx+ctg^*
- j
’ .
22, 13sitU'-12cosx+13sin3;c=30,
n
Javobi: х шап к ± — .
о
Javobi: x
1
n k
> - - —
2
’
23, l+2cos2x+2cos4x+2cos6x®,0.
Javobi: x m—r-
+
x - ± ~ ± ~ .
4
2
3
3
24. cos^+cos
4
(x
”
4
)
,
wlr
Javobi: х = - ^ (k+ l).
217
25.2sin2x+sinx+co&x= 1.
26. 3sinx+sin(x + — ) =1—3 sinxcosx.
Javobi:
(2&+1)
n, keZ.
17-§. Parametrli trigonometrik tenglamalarni yechish
•
Param etrli trigonom etrik tenglam ani yechish param etrlam ing m u m -
kin b o ‘lgan h ar bir qiym atlari sistemasi u ch u n berilgan tenglam aning
h am m a yechim lari to 'p la m in i aniqlash dem akdir. Param etrik k o ‘ri-
nishdagi trigonom etrik tenglam alarni yechish quyidagi ketm a-ketliklar
asosida am alga oshiriladi.
1. P aram etrlam ing m um kin bo'lgan h ar bir qiym atlari sistem asi
u c h u n yechim lar sonini aniqlash.
2. H osil qilingan param etrli yechim form ulalarini topish.
3. T englam ani p aram etrlar asosidagi qiym atlar sistem asini aniqlash.
a
1
-misol. sin(a+x)+sirLx=cos— tenglamani yeching.
Y e c h i s h . Berilgan tenglam aning chap tom onida turgan ifodaga
trigonometrik funksiyalar yig‘indisini ko'paytmaga keltirish formulasini tatbiq
qilsak, u quyidagi ko'rinishni oladi:
27. 3-co s
2
jc—3sinx=0.
и
Javobi: — +2kn, keZ.
_
_
я
28. cos
3
x+4sin
2
2
'Cos
2
j
co&x=0.
Javobi: — +2kn, keZ.
29.
cos
2
2x-sin
2
2 x = - —■
x
x
30. sin
3
.v+2sin 'J cos
2
=0.
x
Javobi: ± ^ + ^ k , keZ.
о
2
Javobi: nk, keZ.
X
31. sinJx cos
2
— sin2xsin
2
= 0
Javobi: nk, keZ.
- . , a
4
a
a
2
sm (— +
x)
cos — = cos —.
2
2
2
Bu tenglamada quyidagi ikki hoi bo‘lishi mumkin.
218
ос
1. Agar c o s - * 0,
a
* (2
n + \)
k
bo'lsa, u holda tenglamani quyidagicha
yozish mumkin:
2 sin(“ +
x)
= 1,
x
=
+ (-1)* ^ +
kn, к
e
Z.
2
2
6
(X
2. Agar
а=(2п+\)к
bo'lsa, u holda cos — = 0 bo'ladi. Bu holda ham
tenglik o'rinli bo'ladi:
Javobi: x
- ~ +
( - l ) fc
+
kn, a
* (
2
« +
l)n,
2
6
-ixtiyoriy
son ,
a = (2 n + l)n.
n
2
-misoI. sin
2
x+asm
22
x=sin — tenglamani yeching.
6
. . .
l - c o s
2
x
2
^ ч
1
Y e c h i s h . ----- ------ +
a(l
- cosz
2x)
= - ,
1
- cos
2
x+
2
a —
2
acos
22
x = l,
2
acos
22
x+cos
2
x-
2
a=
0
,
(cos
2
x) = ~ 1+- ^ + 16— ;
(cos
2
x) =
~ l ~
A -t -l6a—,
4
a
4 a
\
т
—
1
+
"v/l
+
16a2
, ,
I —
1
+
"\/l
+
16a2
I .
a) c o s
2
x = -------- -----------,
l+ 1 6 a2>0,
| --------- -----------|s l,
4a
4a ,
—1+
yji
+ ]бд
2
S4a,
l+ 16a2<4a2+ 8 a+ l,
12
a
2
-
8
a<
0
; a (
12
a -
8)^0
1
) a =
0
,
12
a—
8
*
0
,
2
2
2
) a*
0
,
12
o -
8
=
0
, a = - , bundan a s -
bo'ladi.
0
-
1
- V l + I
6
a
2
, - l - > / l + 16a2
b)
cos
2
x = -------- ---------- ;
| -------------------- |£
1
,
4a
4a
tengsizlik a ning hech qanday qiymatlarida bajarilmaydi. Agar a=0 bo'lsa
f
•
2
О
,
П
,
sin X = - => xw = ± - +
kn, keZ .
219
Javobi:
a*0 da jc,
=izk ±
2
arccos
(
- 1 + Vl + 16a2
4
a
к
4
3-misol.
cosx+sim =a tenglamani yeching.
a
= 0
da jc34= ± y +Atc, &eZ
Y e c h is h . cosx+siiu=
^2
cos(x~ ^ ); agar
\a\< ^2
bo'lsa, bu tenglama
i
7t
d
yechimga ega bo'ladi. Agar
\a\<4l
bo‘lsa, *
=
±
arccos
—
j= + 2kn,
agar
\a\> s/2
bo'lsa, yechim yo'q.
4-misol.
sin2x+3cos2x=a tenglamani yeching.
Y e c h is h . 2smxcosx+3cos
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