Укажите формулу n-ого члена арифметической прогрессии, первые члены которой заданы перечислением: 3, 5,

аn=2n-1,где "n" принадлежит натуральным числам

task/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) .

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) *