Является ли функция f(x) = x-1 унимодальной на [0,2]?

1. 9 —(8x — 11) = 12
9 — 8x + 11 = 12
—8x = 12 — 20 = —8; x = 1;

2. (6x + 1) —(3 — 2x) = 14
6x + 1 — 3 + 2x = 14
6x + 2x = 14 + 2
8x = 16; x = 2;

3. 2x —(6x — 5) = 45
2x — 6x + 5 = 45
2x — 6x = 45 — 5
—4x = 40; x = —10;

4. 5x —(7x + 7) = 9
5x — 7x — 7 = 9
5x — 7x = 9 + 7
—2x = 16; x = —8;

5. 2x —(6x + 1) = 9
2x — 6x — 1 = 9
2x — 6x = 9 + 1
—4x = 10; x = —2,5;

6. 4x —(7x — 2) = 17
4x — 7x + 2 = 17
4x — 7x = 17 — 2
—3x = 15; x = —5; 

7. 2x + 7 = 3x — 2 * (3x — 1)
2x + 7 = 3x — 6x + 2
2x — 3x + 6x = 2 — 7
5x = —5; x = —1;

8. 4 — 2 * (x + 3) = 4 * (x — 5)
4 — 2x — 6 = 4x — 20
—2x — 4x = —20 — 4 + 6
—6x = —18; x = 3;

9. 5x + 3 = 7x — 5 * (2x + 1)
5x + 3 = 7x — 10x — 5
5x — 7x + 10x = —5 — 3
8x = —8; x = —1;

10. 3y —(5 — y) = 11
3y — 5 + y = 11
3y + y = 11 + 5
4y = 16; y = 4.

Решение
1)  2sin x-1=0
sinx = 1/2
x = (-1)^n arcsin(1/2) + πk, k∈Z
x = (-1)^n (π/6) + πk, k∈Z
2)  cos(2x+П/6)+1=0
cos(2x+П/6) = - 1
2x+П/6 = π + 2πn, n∈Z
2x = π - π/6 + 2πn, n∈Z
2x = 5π/6 + 2πn, n∈Z
x = 5π/12 + πn, n∈Z
3)  6sin²x - 5cosx + 5 = 0
6(1 - cos²x) - 5cosx + 5 = 0
6 - 6cos²x - 5cosx + 5 = 0
6cos²x + 5cosx - 11 = 0
cosx = t, ItI ≤ 1
6t² + 5t - 11 = 0
D = 25 + 4*6*11 = 289
t₁ = (- 5 - 17)/12
t₁ = - 22/12
t₁ = -11/6
t₁ = - 1 (5/6) не удовлетворяет условию ItI ≤ 1
t₂ = (- 5 + 11)/12
t₂ = 1/2
cosx = 1/2
x = (+ -)arccos(1/2) + 2πm, m∈Z
 x = (+ -) *(π/3) + 2πm, m∈Z