У=7x-5 => k=7 => f `(x₀)=7
y=x²+6x-8
y`(x)=(x²+6x-8)`=2x+6
y`(x₀)=2x₀+6 и y`(x₀)=k=7
2x₀+6=7
2x₀=1
x₀=0,5 - точка касания
1) pi/2 < a < pi, поэтому sin a > 0, cos a < 0
cos a = -√6/4; cos^2 a = 6/16
sin^2 a = 1 - cos^2 a = 1 - 6/16 = 10/16; sin a = √10/4
tg a = sin a / cos a = (√10/4) : (-√6/4) = -√10/√6 = -√5/√3 = -√15/3
2) 0 < a < pi/2, поэтому sin a > 0, cos a > 0
sin a = √2/3; sin^2 a = 2/9
cos^2 a = 1 - sin^2 a = 1 - 2/9 = 7/9; cos a = √7/3
tg a = sin a / cos a = (√2/3) : (√7/3) = √2/√7 = √14/7
3) 3pi/2 < a < 2pi, поэтому sin a < 0, cos a > 0
cos a = 15/17; cos^2 a = 225/289
sin^2 a = 1 - cos^2 a = 1 - 225/289 = 64/289; sin a = -8/17
tg a = sin a / cos a = (-8/17) : (15/17) = -8/15
7,4a+2,6b-2,5a+3,7b=4,9a+6,3b
2х²-18≥0
2х²≥18
х²≥9
х≤-3
х≥3
Ответ: х£(-ထ;-3]U[3;+ထ).