7cos (2x-π/3) = -3.5|:7
cos (2x-π/3) = -1/2
2x-π/3=+-(π-arccos1/2)+2πn, n∈Z
2x-π/3=+-2π/3+2πn, n∈Z
2x=+-2π/3+π/3+2πn, n∈Z |:2
x=+-π/3+π/6+πn, n∈Z
x1=π/2+πn, n∈Z x2=-π/6+πn, n∈Z
<span>с^2+6с-40=0
c^2+10c-4c-40=0
c(c+10)-4(c+10)=0
(c+10)(c-4)=0
c+10=0
c-4=0
c=-10
c=4</span>
<span><span>(tg40+tg5)/(1-tg40*tg5)=tg(40+5)=tg45=1
</span></span><span />
Cos^2a + sin^2a = 1, значит cos^2a = 1-sin^2a = -(sin^2a-1)
-(sin^2a-1) = 0,3, значит sin^2a-1 = -0,3, то есть 5 sin^2a-1 = -1,5
надеюсь, я правильно прочитал условие задачи
(2x -3)/(5x -20) -(x-2)/(2x-8) =(2x -3)/5(x -4) -(x-2)/2(x-4) = (2(2x-3) -5(x-2) )/10(x-4) =
(4x-6 -5x+10)/10(x-4) = (-x +4)/10(x-4) = -(x-4)/10(x-4) = -1/10 = -0,1.