task/29916604/29916224
1. sin2x = sin(x -π/3) ⇔sin2x + sin(π/3 -x) ⇔2sin(x/2 +π/6)*cos(3x/2 -π/6) =0⇔
[ sin(x/2 +π/6) =0 ; cos(3x/2 -π/6) =0 .⇔ [ x/2 +π/6 =πn ; 3x/2 -π/6 =π/2 + πn , n∈ ℤ .⇔
[ x= - π/3 + 2πn ; x =4π/9 + (2π/3)*n , n∈ ℤ .
2. cos(x - π/6) = cos(π/5) ⇔ cos(x - π/6) - cos(π/5) =0 ⇔
-2sin( (x-π/6-π/5)/2 )*sin( (x-π/6+ π/5)/2) =0⇔ sin( (x-11π/30) /2)*sin((x+π/30)/2)=0 ⇔
[ sin( (x-11π/30) /2) =0 ; sin((x+π/30)/2)=0.⇔[ (x-11π/30)/2 =πn ; (x+π/30)/2=πn , n∈ ℤ ⇔
[ x = 11π/30 +2πn ; x = - π/30 +2πn , n∈ ℤ .
3. cos2x = sin(π/3 +x) ⇔ cos2x = cos(π/2 -(π/3 +x) ) ⇔cos2x - cos(π/6 -x) =0 ⇔
-2sin( (3x -π/6) /2) *sin( ( x +π/6) /2) =0⇔ [sin( (3x -π/6) /2) =0 ;sin( ( x +π/6) /2)=0.⇔
[ ( 3x -π/6)/2 =πn ; (x +π/6)/2 =πn, n∈ ℤ⇔
[ x=π/18+(2π/3)*n ; x = - π/3 +2πn ,n∈ ℤ.
* P.S. sinα+sinβ=2sin((α+β)/2)*cos((α- β)/2) ;cosα-cosβ =-2sin((α -β)/2)*sin((α+β)/2) ; sinα =cos(π/2 - α) *
1) в.
4) а.
5) г.
6) в.
7) - 5,23
8) 6p^2(3p - 2).
9) -7.
10) x = 2.
Объяснение:
7) 4,23a - a^2 = a(4,23 - a) = 5,23(4,23 - 5,23) = 5,23 * (- 1) = - 5,23.
9) 4a - 2(5a - 1) + (8a - 2) = 4a - 10a + 2 + 8a - 2 = 2a = 2 * (- 3,5) = - 7.
10) 5x-9/4 + 5x-7/4 = 1 (у цьому випадку потрібно домножити на 4)
4(5x - 9)/4 + 4(5x - 7)/4 = 1 * 4 (тоді 4 скоротяться і не буде знаменника)
5x - 9 + 5x - 7 = 4
5x + 5x = 4 + 9 + 7 (при перенесенні через = знак змінюється)
10x = 20
x = 20 / 10
x = 2