1
8(√3/2*cosx-1/2*sinx)-4cosx(√3/2cosx-1/2*sinx)=0
cos(x+π/6)*(8-4cosx)=0
cos(x+π/6)=0⇒x+π/6=π/2+πn,n∈z⇒x=π/3+πn,n∈z
8-4cosx=0⇒cosx=2>1 нет решения
2
4sin²3x+4cos²3x-cos²3x-3sin²3x-6sin3xcos3x=0/cos²3x
tg²3x-6tgx+3=0
tg3x=a
a²-6a+3=0
D=36-12=24
a1=(6-2√6)/2=3-√6⇒tg3x=3-√6⇒3x=arctg(3-√6)+πn,n⇒z⇒
x=1/3*arctg(3-√6)+πn/3,n∈z
a2=(6+2√6)/2=3+√6⇒tg3x=3+√6⇒3x=arctg(3+√6)+πk,k⇒z⇒
<span>x=1/3*arctg(3+√6)+πk/3,k∈z
3
8(</span>√3/2sinx+1/2cosx)-4sinx(√3/2sinx+1/2cosx)=0
cos(x-π/6)*(8-4sinx)=0
<span>cos(x-π/6)=0⇒x-π/6=π/2+πn,n∈z⇒x=2π/3+πn,n∈z
</span>8-4sinx=0⇒sinx=2>1 нет решения
4
1)sinx>0⇒x∈(2πn;π+2πn,n∈z)
sinx/sinx-2=2cosx
2cosx=-1
cosx=-1/2
x=2π/3+2πn,n∈z
2)sinx≤0⇒x∈[π+2πn;2π+2πn,n∈z]
-sinx/sinx-2=2cosx
2cosx=-3
cosx=-1,5<-1 нет решения
1)8x²+2px-3p²;2x²+3px-3p²;
сумма:(8x²+2px-3p²)+(2x²+3px-3p²)=8x²+2px-3p²+2x²+3px-3p²=10x²+5px-6p²;
разность:(8x²+2px-3p²)-(2x²+3px-3p²)=8x²+2px-3p²-2x²-3px+3p²=6x²-px=x(6x-p);
2)(3a+5b)-(9a-7b)+(-5a+11b)=3a+5b-9a+7b-5a+11b=-11a+23b;
(3x²+2x)+(2x²-3x-4)-(-x²+19)=3x²+2x+2x²-3x-4+x²-19=6x²-x-23;
Решение
1) 4sin130° / [sin65° * sin25°] =
= [4*sin(180° - 50°)] / [1/2 * (cos(65° - 25) - cos(65° + 25°)] =
= 8*sin50° / [cos40° - cos90°] = [8*sin(90° - 40°)] / cos40° =
= 8*cos40° / cos40° = 8
1) 3/5x - 1/2x = 0,6x - 0,5x = 0,1x
2) x = - 10
0,1 * ( - 10 ) = - 1
Ответ минус 1
