6sin x = 6 * 1/2 = 3
2cos 6x = 2 * (-1) = -2
sin 3x = sin 90 = 1
3 - (-2) + 1 = 6
1.149. ((x+3)/(x²-3x)+(x-3)/(x²+3x))·(9x-x³)/(x²+9)=
=((x+3)/x(x-3)+(x-3)/x(x+3))·x(9-x²)/(x²+9)=[((x+3)²+(x-3)²)/x(x²-9)]·x(9-x²)/(x²+9)=
=(x²+6x+9+x²-6x+9)·x·(9-x²)/(x(x²-9)·(x²+9)=(2·(x²+9)·x(9-x²))/(x·(x²-9)·(x²+9))=-2;
1.150 [(x+3)/(x-3)-(x-3)/(x+3)]:2x/(9-x²)=((x+3)²-(x-3)²)/(x²-9):2x/(9-x²)=
=(x²+6x+9-x²+6x-9)·(9-x²)/(2x·(x²-9))=(12x(9-x²)/2x(x²-9)=-6;
1.151 2a/(a+1)+(3/(a-1)²-3/(a²-1)):3/(a²-2a+1)=
=2a/(a+1)+[(3·(a+1)-3(a-1))/(a-1)²(a+1)]:3/(a-1)²=
=2a/(a+1)+(3a+3-3a+3)·(a-1)²/[3(a-1)²·(a+1)]=2a/(a+1)+6/3(a+1)=(2a+2)/(a+1)=2;
5Cos²x - 3Cosx -2 = 0
D = 9 + 40 = 49
a) Cosx = 1 б) Cosx = -0,4
x=2πk , k ∈Z x = +-arcCos(-0,2) + 2πn , n ∈Z
2) Sin²x - 6Sinx = 0
Sinx(Sinx - 6) = 0
Sinx = 0 или Sinx -6 = 0
x = πn , n ∈Z Sinx = 6
∅
3) 3Sinx -2Cos²x = 0
3Sinx -2(1 - Sin²x) = 0
3Sinx -2 + 2Sin²x = 0
2Sin²x + 3Sinx -2 = 0
D = 9 + 16 = 25
a) Sinx = 1/2 б) Sinx = -2
x = (-1)ⁿ π/6 + πn , n ∈Z ∅
4) Sin4xCos2x - Sin2xCos4x = 0
Sin2x = 0
2x = πn , n ∈Z
x = πn/2 , n ∈Z
5) Sin²x + 2SinxCosx + Cos²x = 1 + SinxCosx
SinxCosx = 0
Sinx = 0 или Cosx = 0
x = nπ, n ∈Z x = π/2 + πk , k ∈Z
Делим 8,5 на 1,7, и получаем 5, затем от 5 вычитаем 1,2 = 3,8.
Иначе: 1. 8,5 : 1,7 = 5
2. 5-1,2=3,8
<span>Теорему синусов можно записать в виде
</span>
<span>Пользуясь этой формулой, найдите ?, если ? = 15, sin ?= 1/5 и sin? = 1/ 4
Предположим такое условие:
1) найдите а, </span>если b= 15, sin a= 1/5 и sin b = 1/ 4