Решите неравенство: sin^6x+cos^6x>5/8​

1) 7tgx -10Ctgx +9 =0 |* tgx
     7tg^2x -10 +9tgx = 0
tgx = y
7y^2 +9y -10 = 0
y1 = 10/14 = 5/7
у2 = -2
а) у = 5/7
tgx = 5/7
x = arctg5/7 + k, k ЄZ
б) у = -2
tgx = -2
x = -arctg2 +k, k ЄZ
2) 10SinxCosx -14Cos^2x +2*1 = 0
10SinxCosx -14Cos^2x +2(Sin^2x+Cos^2x) = 0
10SinxCosx -14Cos^2x +2Sin^2x +2Cos^2x = 0
10SinxCosx -12Cos^2x +2Sin^2x = 0 :Cos^2x
10tgx -12 +2tg^2x= 0
tgx = y
2y^2 +10y -12=0
y^2 + 5y - 6 = 0
По т. Виета у1 = - 6 и  у2 = 1
а) у = - 6
tgx = -6
x = -arctg6+k, kЄZ 
б)у = 1
tgx = 1
x =  + 
3) 9(Cos^2x - Sin^2x) -4Cos^2x = 22SinxCosx + 9*1
9Cos^2x - 9Sin^2x -4Cos^2x -22SinxCosx -9(Sin^2x+Cos^2x) = 0
9Cos^2x - 9Sin^2x -4Cos^2x -22SinxCosx -9Sin^2x - 9Cos^2x = 0
-18Sin^2x -4Cos^2x -22SinxCosx = 0
9Sin^2x +2Cos^2x +11SinxCosx = 0|:Cos^2x
9tg^2x +2 +11tgx = 0
tgx = y
9y^2 +11y +2 = 0
y1=-1,  y2 = -2/9
a) y = -1
tgx = -1
x = - + 
б) у = -2/9
tgx = -2/9
x = -arctg(2/9) + 
 

-

1. 3х - 3

2. -11

3. 7х - 1

4. -20

5. 5

6. 2х - 9

7. 2

8. 7х - 10

9. -19

10. 7х - 5

Объяснение:

1. 3(х+4) - (3-х) - х - 4 = 3х + 4 - 3 + х - х - 4 = 3х - 3

2. x + 4 - 5(2-х) - (5+1)х - 5 = х + 4 - 10 + 5х - 5х - х - 5 = 4 - 10 - 5 = -11

3. 4(x+4) - 4(3-х) - x - 5 = 4х + 16 - 12 + 4х - х - 5 = 4х + 4х - х + 16 - 12 - 5 = 7х - 1

4. x + 2 - 4(5-х) - (4+1)х - 2 = х + 2 - 20 + 4х - 4х - х - 2 = -20

5. 2(x+4) - (1-x) - (1+2)х - 2 = 2х + 8 - 1 + х - х - 2х - 2 = 8 - 1 - 2 = 5

6. x + 2 - 2(5-х) - x - 1 = х + 2 - 10 + 2х - х - 1 = 2 - 10 - 1 + 2х = -9 + 2х = 2х - 9

7. 4(x+2) - (1-x) - (1+4)х - 5 = 4х + 8 - 1 + х - х - 4х - 5 = 8 - 1 - 5 = 2

8. 4(х+2) - 4(4-x) - x - 2 = 4х + 8 - 16 + 4х - х - 2 = 4х + 4х - х + 8 - 16 - 2 = 7х - 10

9. 3(х+1) - 4(5-х) - (4+3)х - 2 = 3х + 3 - 20 + 4х - 4х - 3х - 2 = 3 - 20 - 2 = -19

10. 3(x+3) - 5(2-х) - x - 4 = 3х + 9 - 10 + 5х - х - 4 = 3х + 5х - х + 9 - 10 - 4 = 7х - 5