Question

Решите тригонометрическое уравнение:
1) cos2x+2sin2x+2=0
2) 2cos2x+2sin^2x=5+4sin2x
3) 2sin2x=3-2sin^2x

Answer 1

1) cos2x+2sin2x+2=0

(1 - tg²x)/(1 + tg²x) + 2*2tgx/(1 + tg²x) +2 = 0 | * (1 + tg²x )≠ 0

1 - tg²x + 4tgx +2(1 +tg²x) = 0

1 - tg²x + 4tgx +2 +2tg²x = 0

tg²x +4tgx +3 = 0

По т. Виета корни -1 и -3

а) tgx = -1 б) tgx = -3

x = -π/4 + πk, k ∈Z x = -arctg 3 + πn , n ∈Z

2) 2cos2x+2sin²x=5+4sin2x

2(Cos²x - Sin²x) +2Sin²x = 5 +8SinxCosx

2Cos²x - 2Sin²x +2Sin²x = 5*1 +8SinxCosx

2Cos²x = 5*(Sin²x + Cos²x) + 10SinxCosx

2Cos²x = 5Sin²x + 5Cos²x +8SinxCosx

5Sin²x +3Cos²x +8sinxCosx = 0 |: Cos²x

5tg²x + 8tgx +3 = 0

tgx = t

5t² +8t +3 = 0

t = (-4 +-√(16 -15))/5 = (-5 +-1)/5

t₁ = -6/5= -1.2 t₂ = -4/5= -0,8

tgx =- 1,2 tgx = -0,8

x = -arctg1,2+ πk , k ∈ Z x = -arctg0,8 + πn , n ∈Z

3) 2sin2x=3-2sin²x

6SinxCosx = 3Sin²x + 3Cos²x -2Sin²x

6SinxCosx = Sin²x + 3Cos²x | : Cos²x

6tgx = tg²x +3

tgx = t

t² -6t +3 = 0

t =3+-√(9 -3)

t₁ = 3 +√6 t₂ = 3 - √6

tgx = 3 +√6 tgx = 3 - √6

x = arctg(3 + √6) + πk , k ∈Z x = arctg(3 - √6) + πn , n ∈ Z

Answer 2

тогда t=1;3

Answer 3

Да, кстати, 2sin2x = 4sinxcosx.

Answer 4

Да, ошиблась. 2sin2x = 4sinxcosx

Answer 5

4tgx = tg²x +3, t² -4t +3 = 0, корни 1 и 3, ответ: х = π/4 + πk, k ∈ Z

Answer 6

x = arctg3 + πk, k ∈ Z.