Очень прошу,помогите! Алгебра,10 класс
1)2sin²x=√3cos(П/2+x), [3П/2;3П]
2)2sin³x=cos(x-П/2), [-3П/2;-П/2]
1)2sin²x=√3cos(П/2+x), [3П/2;3П]
2)2sin³x=cos(x-П/2), [-3П/2;-П/2]
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Answers & Comments
sinx(sinx+ √3/2) =0
a) sinx=0 ⇒ x = πk ; x∈[3π/2;3π] ⇒ x1=2π; x2 = 3π
b) sinx = - √3/2 ⇒ x= - π/3 ·(-1)^k +πk ; ⇒ x = -π/3+2π=5/3π
Ответ: x = {5/3π ; 2π ; 3π} = { 300° ; 360° ; 540°}
2) 2sin³x= cos(x-π/2)=cos(π/2-x)= sinx
sinx(sin²x - 1/2)=0
a) sinx=o ⇒ x = πk ; x∈[-3π/2; - π/2] ⇒ x = -π
b) sin²x = 1/2 ⇒ C) sinx = -√2/2 .; x = -π/4·(-1)^k+πk ⇒ x=-3/4π
D) sinx= √2/2 ; x = π/4·(-1)^k+ πk ⇒ x= -5/4π
Ответ: x = { -3/4π ; -π ; - 5/4π }