Решите неравенство:
cosx≤2+x^4

1) Если x < 3, то |x - 3| = 3 - x
10*|3 - x - 4| = x + 2
10*|-x - 1| = x + 2
10*|x + 1| = x + 2

1а) Если x < -1, то |x + 1| = -x - 1
10(-x - 1) = x + 2
-10x - 10 = x + 2
-12 = 11x
x1 = -12/11 < -1 - подходит

1б) Если -1 <= x < 3, то |x + 1| = x + 1
10(x + 1) = x + 2
10x + 10 = x + 2
9x = -8
x2 = -8/9 > -1 - подходит

2) Если x >= 3, то |x - 3| = x - 3
10*|x - 3 - 4| = x + 2
10*|x - 7| = x + 2

2а) Если 3 <= x < 7, то |x - 7| = 7 - x
10(7 - x) = x + 2
70 - 10x = x + 2
11x = 68
x3 = 68/11 < 7 - подходит

2б) Если x >= 7, то |x - 7| = x - 7
10(x - 7) = x + 2
10x - 70 = x + 2
9x = 72
x4 = 8 > 7 - подходит.
ответ: -12/11; -8/9; 68/11; 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) + 
 

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