Решите тригонометрическое уравнение

1) 2sin^2x+3sinx-5=0

2) 10sin^2x-17cosx-16=0

3) 5sin^2x+1&sinxcosx+6cos^2x=0

4) 3tgx-14ctgx+1=0

1) 2sin²x + 3sinx - 5 = 0

Пусть t = sinx, t ∈ [-1; 1].

2t² + 3t - 5 = 0

D = 9 + 4•5•2 = 49 = 7²

t1 = (-3 + 7) / 4 = 4/4 = 1

t2 = (-3 - 7) / 4 = - 10/4 - не уд. условию

Обратная замена:

sinx = 1

x = π/2 + 2πn, n ∈ Z.

2) 10sin²x - 17cosx - 16 = 0

10 - 10cos²x - 17cosx - 16 = 0

-10cos²x - 17cosx - 6 = 0

10cos²x + 17cosx + 6 = 0

Пусть t = cosx, x ∈ [-1; 1].

D = 289 - 4•6•10 = 49 = 7²

t1 = (-17 + 7) / 20 = - 10/20 = - 1/2

t2 = (-17 - 7) / 20 = - 24/20 - не уд. условию

Обратная замена:

cosx = - 1/2

x = ±arccos (-1/2) + 2πn, n ∈ Z

x = ±2π/3 + 2πn, n ∈ Z.

3) 5sin²x + 17sinxcosx + 6cos²x = 0

Разделим на cos²x.

5tg²x + 17tgx + 6 = 0

Пусть t = tgx.

D = 289 - 6•4•5 = 289 - 120 = 13²

t1 = (-17 + 13) / 10 = - 4/10 = - 2/5

t2 = (-17 - 13) / 10 = - 30/10 = - 3

Обратная замена:

tgx = - 2/5

x = arctg (-2/5) + πn, n ∈ Z.

x = arctg (-3) + πn, n ∈ Z.

4) 3tgx - 14ctg + 1 = 0

3tgx - 14/tgx + 1 = 0

3tg²x + tgx - 14 = 0

Пусть t = tgx.

3t² + t - 14 = 0

D = 1 + 14•4•3 = 13²

t1 = (-1 + 13) / 6 = 12/6 = 2

t2 = (-1 - 13) / 6 = - 14/6 = - 7/3

обратная замена:

tgx = 2

x = arctg2 + πn, n ∈ Z

tgx = - 7/3

x = arctg (-7/3) + πn, n ∈ Z.