Trigonometrie plana si sferica Ecuatii trigonometrice

1 Exercitii rezolvate

1. cos 2x + 4 sinx 1 = 0R: Inlocuim n ecuatie cos 2x = 1 2 sin2 x si obtinem

1 2 sin2 x + 4 sinx 1 = 0 2 sinx(2 sinx) = 0Cum sinx 1 2 sinx 0 deci singura solutie acceptabila este sinx = 0 de unde obtinemx = kpi + (1)k arcsin 0, k Z x = kpi, k Z

2.

2 cosx + 2 sin2 x + ctg2 x 3 = 0, x kpi, k ZR: Folosind formulele care exprima sin2 x si ctg2 x n functie de cos2 x obtinem

2 cosx + 2(1 cos2 x) + cos2 x1 cos2 x 3 = 0

Punand t = cosx, n urma calculelor se obtine2t4 2t3 +2t 1 = 0 (2t 1)(2t3 + 1) = 0

t1 = 12= 2

2 x = 2kpi pi

4; t2 = 162 x = 2kpi arccos( 162)

3. 4 sinx + 2 cosx 3 tgx 2 = 0R: Facem substitutia t = tg x

2. Avem sinx = 2t

1 + t2 , cosx = 1 t21 + t2 si tgx = 2t1 t2 . Dupaefectuarea calculelor se obtine ecuatia

2t4 7t3 2t2 + t = 0care are radacinile t1 = 0, t2 = 12 , t3,4 = 2 3.

tgx

2= 0 x

2= kpi x = 2kpi, k Z

tgx

2= 1

2 x

2= kpi + arctg (1

2) x = 2kpi + 2 arctg (1

2) , k Z

tgx

2= 2 +3 x = 2kpi + 2 arctg (2 +3) , k Z

tgx

2= 2 3 x = 2kpi + 2 arctg (2 3) , k Z

4.

3 sinx + cosx = 1R: Impartind prin

3 obtinem sinx+ 1

3cosx = 1

3. Punem 1

3= tg pi

6 sinx+ sin pi6

cos pi6cosx =

13 sinx cos pi

6+ cosx sin pi

6= 1

3cos pi

6de unde folosind formula pentru sinusul sumei gasim

sin (x + pi6) = 1

2, asadar

x + pi6= kpi + (1)k arcsin 1

2 x = kpi + ((1)k 1) pi

6

5. 3 sin2 x + 2 sinx cosx cos2 x = 0R: Impartind prin cos2 x 0 si punand t = tgx obtinem ecuatia

3t2 + 2t 1 = 0t1 = 1 tgx = 1 x = kpi + arctg(1) = kpi pi4 , k Zt2 = 13 tgx = 13 x = kpi + arctg ( 13) , k Z

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Trigonometrie plana si sferica Ecuatii trigonometrice

6. 2 sin4 x 23 sin3 x cosx + 4 sin2 x cos2 x + 23 sinx cos3 x 2 cos4 x = 1R: Inlocuind n membrul drept 1 = sin2 x + cos2 x ecuatia devine:

2 sin4 x 23 sin3 x cosx + 4 sin2 x cos2 x + 23 sinx cos3 x 2 cos4 x = (sin2 x + cos2 x)2sin4 x 23 sin3 x cosx + 2 sin2 x cos2 x + 23 sinx cos3 x 3 cos4 x = 0

Impartind prin cos4 x si notand t = tgx obtinemt4 23t3 + 2t2 + 23 3 = 0 (t2 1)(t2 23 + 3) = 0

t1,2 = 1 tgx = 1 x = kpi pi4 , k Zt3,4 =3 tgx =3 x = kpi + pi3 , k Z

7. 5(sinx + cosx) 2 sin 2x = 4R: Facem substitutia u = sinx + cosx sin 2x = u2 1. Se obtine ecuatia 2u2 5u + 2 = 0 curadacinile reale u1 = 2, u2 = 12 .u = sinx + cosx =2 cos (x pi

4) = 2 cos (x pi

4) =2 > 1 nu exista solutii;

u = sinx + cosx =2 cos (x pi4) = 1

2 cos (x pi

4) = 2

4 x = 2kpi arccos 2

4+ pi

4.

8. cos2 x + cos2 2x cos2 3x = 1R: 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 diferentei si sumei n produs gasim 2 sin 4x sin 2x = 2 sin2 2x2 sin 2x(sin 4x + sin 2x) = 0 4 sin 2x sin 3x cosx = 0sin 2x = 0 2x = kpi x = kpi

2, k Z

sin 3x = 0 3x = kpi x = kpi3, k Z

cosx = 0 x = (2k + 1)pi2, k Z, multime de solutii care este inclusa n prima multime.

9. 2(sin6 x + cos6 x) + sin4 x + cos4 x = 1R: Cu substitutia y = sin 2x avem sin4 x+cos4 x = 1 1

2y2, sin6 x+cos6 x = 1 3

4y2, iar ecuatia

devine 2 (1 34y2) + 1 1

2y2 = 1 y2 = 1 cu radacinile y = 1.

sin 2x = 1 2x = 2kpi + pi2 x = kpi + pi

4, k Z;

sin 2x = 1 2x = 2kpi pi2 x = kpi pi

4, k Z.

10. cosx cos 7x = cos 3x cos 5xR: Transformand cele doua produse n sume avem

1

2(cos 8x + cos 6x) = 1

2(cos 8x + cos 2x) cos 6x cos 2x = 0 2 sin 4x sin 2x = 0

sin 4x = 0 x = kpi4, k Z

sin 2x = 0 x = kpi2, k Z, multime de solutii care este inclusa n prima multime.

2 Tema

1. cosx +3 sinx =m; discutie dupa m R2. 2 cos2 x sin 2x + sinx + cosx = 13. cos2 x + 3 sin2 x + 23 sinx cosx = 14. cos2 x + cos2 2x + cos2 3x + cos2 4x = 25. sin3 x cos 3x + sin 3x cos3 x = 3

4

2