Решите тригонометрические уравнения:

1. 2cos^2x+3cosx-5=0

2. 6cos^2x-11sinx-10=0

3. Sin^2x+7sinx cosx+12cos^2x=0

4. 7tgx-8ctgx+10=0

5. 9cos^2x-sin^2x=8sinx cosx

1. 2cos²x + 3cosx - 5 = 0

Пусть t = cosx, t ∈ [-1; 1].

2t² + 3t - 5 = 0

D = 9 + 2•4•5 = 49 = 7²

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

t2 = (-3 - 7) / 4 = - 10/4 = - 2,5 - не уд. условию.

Обратная замера:

cosx = 1

x = 2πn, n ∈ Z

2. 6cos²x - 11sinx - 10 = 0

6 - 6sin²x - 11sinx - 10 = 0

-6sin²x - 11sinx - 4 = 0

6sin²x + 11sinx + 4 = 0

Пусть t = sinx, t ∈ Z.

6t² + 11t + 4 = 0

D = 121 - 4•6•4 = 25 = 5²

t1 = (-11 + 5) / 12 = - 1/2

t2 = (-11 - 5) / 12 = - 16/12 - не уд. условию.

Обратная замена:

sinx = - 1/2

x = (-1) ⁿ+¹π/6 + πn, n ∈ Z.

3. sin²x + 7sinxcosx + 12cos²x = 0

tg²x + 7tgx + 12 = 0

Пусть t = tgx.

t² + 7t + 12 = 0

D = 49 - 48 = 1

t1 = (-7 + 1) / 2 = - 6/2 = - 3

t2 = (-7 - 1) / 2 = - 8/2 = - 4

Обратная замена:

tgx = - 3

x = arctg (-3) + πn, n ∈ Z

tgx = - 4

x = arctg (-4) + πn, n ∈ Z.

4. 7tgx - 8ctgx + 10 = 0

7tgx - 8/tgx + 10 = 0

7tg²x + 10tgx - 8 = 0 (tgx ≠ 0)

Пусть t = tgx.

7t² + 10t - 8 = 0

D = 100 + 4•7•8 = 324 = 18²

t1 = (-10 + 18) / 14 = 8/14 = 4/7

t2 = (-10 - 18) / 14 = - 28/14 = - 2

Обратная замена:

tgx = 4/7

x = arctg4/7 + πn, n ∈ Z

tgx = - 2

x = arctg (-2) + πn, n ∈ Z.

5. 9cos²x - sin²x = 8sinxcosx

9 - tg²x = 8tgx

tg²x + 8tgx - 9 = 0

Пусть t = tgx.

t² + 8t - 9 = 0

t1 + t2 = - 8

t1•t2 = - 9

t1 = - 9

t2 = 1

Обратная замена:

tgx = - 9

x = arctg (-9) + πn, n ∈ Z.

tgx = 1

x = π/4 + πn, n ∈Z.