Question

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

Answer 1

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.