Task/25864040
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решить УРАВНЕНИЕ cosx*cos2x*sin3x = (1/4)sin2x
sin3x*cosx*cos2x = (1/4)sin2x ;
(1/2)(sin4x +sin2x) * cos2x =(1/4)sin2x ;
(1/2)sin4cos2x +(1/2)sin2x*cos2x = (1/4)sin2x ;
(1/4)sin6x +(1/4)sin2x +(1/4)sin4x = (1/4)sin2x ;
(1/4) (sin6x +sin4x) =0 ;
(1/2)(sin5x*cosx =0 ;
[ sin5x = 0 ; [ 5x =π*n ; [ x = (π/5)*n ;
[ cosx = 0 . [ x = π/2 +π*n. [ x = π/2 +π*n. n ∈ Z.
ответ : (π/5)*n , π/2 +π*n. n ∈ Z.
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sinα*cosβ = (1/2)* (sin(α+β) +sin(α- β) ) ;
sin2α =2sinα*cosα ⇔sinα*cosα = (sin2α) / 2.