<span>y=x-1/x
x`=1
(1/x)`=1/x*1/x=1/x</span>²
y`=1+1/x²
1) cos(x)^2 - sin(x)^2 = -1
1-sin(x)^2 - sin(x)^2 = -1
2sin(x)^2 = 2
x = (-1)^n * p/2 + n*p
2) 2cos(x)*sin(x) = 0, значит либо sin(x) = 0, либо cos(x) = 0
sin(x) = 0 x = 0 + n*p
cos(x) = 0 x = p/2 + n*p
3) 1 + cos(6x) = cos(3x)
1 + cos(3x)^2 - sin(3x)^2 - cos(3x) = 0
2cos(3x)^2 - cos(3x) = 0
2cos(3x)*(cos(3x)-1/2) = 0
cos(3x) = 0 x = p/2 + n*p
cos(3x) = 1/2 x = p/9 + n*p/3
Log₃(7-2x)=2
7-2x>0
2x<7
x<3,5
log₃(7-2x)=log₃9
7-2x=9
2x=7-9
2x=-2
x=-1 (принадлежит интервалу области определения (-∞;3,5)
Ответ:-1