1 Exercit˘ii rezolvate - Deliu
[Pages:2]Trigonometrie plana ?si sferica
Ecua?tii trigonometrice
1 Exerci?tii rezolvate
1. cos 2x + 4 sin x - 1 = 0 R: ^Inlocuim ^in ecua?tie cos 2x = 1 - 2 sin2 x ?si ob?tinem
1 - 2 sin2 x + 4 sin x - 1 = 0 2 sin x(2 - sin x) = 0
Cum sin x 1 2 - sin x 0 deci singura solu?tie acceptabila este sin x = 0 de unde ob?tinem
x = k + (-1)k arcsin 0, k Z x = k, k Z 2. 2 cos x + 2 sin2 x + ctg2 x - 3 = 0, x k, k Z R: Folosind formulele care exprima sin2 x ?si ctg2 x ^in func?tie de cos2 x ob?tinem
2
cos
x
+
2(1
-
cos2
x)
+
1
cos2 x - cos2
x
-
3
=
0
Pun^and t = cos x, ^in urma calculelor se ob?tine
2t4 - 2t3 + 2t - 1 = 0 ( 2t - 1)( 2t3 + 1) = 0
t1 = 1 = 2
2 2
x
=
2k
?
;
4
t2
=
-
1 62
x
=
2k
?
arccos
- 1 62
3. 4 sin x + 2 cos x - 3 tg x - 2 = 0
R:
Facem
substitu?tia
t
=
tg
x 2
.
Avem
sin x
=
2t 1 + t2 ,
cos x
=
1 - t2 1 + t2
?si
tg x
=
2t 1 - t2 .
Dupa
efectuarea calculelor se ob?tine ecua?tia
2t4 - 7t3 - 2t2 + t = 0
care
are
radacinile
t1
=
0,
t2
=
-
1 2
,
t3,4
=
2
?
3.
x tg
2
=0
x 2
= k x = 2k,
kZ
tg x = - 1 x = k + arctg - 1 x = 2k + 2 arctg
tg
2 x
2 2 =2+ 3
x
=
2k
2 + 2 arctg
tg
2 x
=2- 3
x
=
2k
+ 2 arctg
2
2+ 3
2- 3
, ,
kZ kZ
-1 2
, kZ
4.
3 sin x + cos R: ^Impar?tind
x=1 prin
3
ob?tinem sin x + 1 cos x =
3
1 .
3
Punem
1 3
=
tg
6
sin
x
+
sin cos
6 6
cos x =
1 3
sin
x
cos
6
+
cos
x
sin
6
=
1 3
cos
6
de
unde
folosind
formula
pentru
sinusul
sumei
gasim
sin
x
+
6
=
1 2
,
a?sadar
x + = k + (-1)k arcsin 1 x = k + (-1)k - 1
6
2
6
5. 3 sin2 x + 2 sin x cos x - cos2 x = 0 R: ^Impar?tind prin cos2 x 0 ?si pun^and t = tg x ob?tinem ecua?tia
3t2 + 2t - 1 = 0
t1
= -1 tg x = -1 x = k + arctg(-1) = k -
4
,
kZ
t2
=
1 3
tg x
=
1 3
x
=
k
+
arctg
1 3
, kZ
1
Trigonometrie plana ?si sferica
Ecua?tii trigonometrice
6. 2 sin4 x - 2 3 sin3 x cos x + 4 sin2 x cos2 x + 2 3 sin x cos3 x - 2 cos4 x = 1
R: ^Inlocuind ^in membrul drept 1 = sin2 x + cos2 x ecua?tia devine:
2 sin4 x - 2 3 sin3 x cos x + 4 sin2 x cos2 x + 2 3 sin x cos3 x - 2 cos4 x = (sin2 x + cos2 x)2
sin4 x - 2 3 sin3 x cos x + 2 sin2 x cos2 x + 2 3 sin x cos3 x - 3 cos4 x = 0
^Impar?tind prin cos4 x ?si not^and t = tg x ob?tinem
t4 - 2 3t3 + 2t2 + 2 3 - 3 = 0 (t2 - 1)(t2 - 2 3 + 3) = 0
t1,2
= ?1 tg x = ?1 x = k ?
4
,
kZ
t3,4 =
3 tg x =
3 x = k +
3
,
kZ
7. 5(sin x + cos x) - 2 sin 2x = 4
R: Facem substitu?tia u = sin x + cos x sin 2x = u2 - 1. Se ob?tine ecua?tia 2u2 - 5u + 2 = 0 cu
radacinile
reale
u1= 2,
u2 =
1 2
.
u = sin x + cos x = 2 cos
x
-
4
u = sin x + cos x =
2 cos
x
-
4
= 2 cos
x
-
4
=
1 2
cos
x
-
4
= 2 > 1 nu exista solu?tii;
=
2 4
x
=
2k
?
arccos
2 4
+
4
.
8. cos2 x + cos2 2x - cos2 3x = 1
R:
1 2
(1
+
cos 2x)
+
1 2
(1
+
cos 4x)
-
1 2
(1
+
cos 6x)
=
1
cos 2x
-
cos
6x
=
1
-
cos 4x.
Folosind
formulele de transformare a diferen?tei ?si sumei ^in produs gasim -2 sin 4x sin 2x = 2 sin2 2x
2 sin 2x(sin 4x + sin 2x) = 0 4 sin 2x sin 3x cos x = 0
sin 2x = 0 2x = k
x=
k 2
,
k
Z
sin 3x = 0 3x cos x = 0 x =
= k (2k + 1)
x
2
= ,
k
k3
,k Z,
Z mul?time
de
solu?tii
care
este
inclusa
^in
prima
mul?time.
9. 2(sin6 x + cos6 x) + sin4 x + cos4 x = 1
R:
Cu
substitu?tia
y
= sin 2x
avem
sin4
x
+
cos4
x
=
1
-
1 2
y2,
sin6
x
+
cos6
x
=
1
-
3 4
y2,
iar
ecua?tia
devine 2 sin 2x = 1
1 -243xy=2
sin 2x = -1 2x
=2+k2k1+--212
y2 = 1 y2 = 2xx==kk+-4
1 cu
, k
4
,
k
radacinile
Z; Z.
y
=
?1.
10. cos x cos 7x = cos 3x cos 5x R: Transform^and cele doua produse ^in sume avem
1 (cos 8x + cos 6x) = 1 (cos 8x + cos 2x) cos 6x - cos 2x = 0 -2 sin 4x sin 2x = 0
2
2
sin 4x = 0 x =
k 4
,
k
Z
sin 2x = 0 x =
k 2
,
k
Z,
mul?time
de
solu?tii
care
este
inclusa ^in
prima
mul?time.
2 Tema
1. cos x + 3 sin x = m; discu?tie dupa m R
2. 2 cos2 x - sin 2x + sin x + cos x = 1
3. cos2 x + 3 sin2 x + 2 3 sin x cos x = 1
4. cos2 x + cos2 2x + cos2 3x + cos2 4x = 2
5.
sin3 x cos 3x + sin 3x cos3 x =
3 4
2
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