Решить уравнение sin2x + cos2x = √2sin3x
ответ: x =π/4 + 2π*k , k ∈ ℤ ;
x = 3π/20 + (2π/5)*n , n ∈ ℤ .
Объяснение:
sin2x+cos2x=√2sin3x ⇔(1/√2)*sin2x+(1/√2)*cos2x=sin3x ⇔
cos(π/4)*sin2x+sin(π/4)*cos2x = sin3x ⇔ sin(2x+π/4) = sin3x ⇔
sin3x - sin(2x+π/4) = 0 ⇔ 2sin( (x - π/4) / 2 ) *cos( (5x +π/4) /2 )= 0 ⇔
a) (x -π/4) / 2 =π*k , k ∈ ℤ ⇒ x =π/4 + 2π*k , k ∈ ℤ ;
b) (5x +π/4) / 2 = π/2+ π*n , n ∈ ℤ ⇒ 5x +π/4 = π+2π*n , n ∈ ℤ ⇔
* * * a*sinx +b*cosx = √(a² +b²) sin(x+φ) , где φ=arctg(b/a) * * *
Решить уравнение sin2x + cos2x = √2sin3x
ответ: x =π/4 + 2π*k , k ∈ ℤ ;
x = 3π/20 + (2π/5)*n , n ∈ ℤ .
Объяснение:
sin2x+cos2x=√2sin3x ⇔(1/√2)*sin2x+(1/√2)*cos2x=sin3x ⇔
cos(π/4)*sin2x+sin(π/4)*cos2x = sin3x ⇔ sin(2x+π/4) = sin3x ⇔
sin3x - sin(2x+π/4) = 0 ⇔ 2sin( (x - π/4) / 2 ) *cos( (5x +π/4) /2 )= 0 ⇔
a) (x -π/4) / 2 =π*k , k ∈ ℤ ⇒ x =π/4 + 2π*k , k ∈ ℤ ;
b) (5x +π/4) / 2 = π/2+ π*n , n ∈ ℤ ⇒ 5x +π/4 = π+2π*n , n ∈ ℤ ⇔
x = 3π/20 + (2π/5)*n , n ∈ ℤ .
* * * a*sinx +b*cosx = √(a² +b²) sin(x+φ) , где φ=arctg(b/a) * * *