Ответы на вопрос:
Task/27400429 решите sin(x+30)+cos(x+60 ) =1+cos2x cos(x+60°)+sin(x+30°) =1+cos2x ; 1 способ cosx*cos60° - sinx*sin60° +sinx*cos30° +cosx*sin30° =1+cos2x ; (1/2)*cosx - (√3 /2 )sinx + sinx* (√3 /2 ) +cosx*(1/2) =2cos²x ; cosx = 2cos²x ; 2cosx (cosx -1/2)= 0 ; cosx =0 ⇒ x =π/2+πn , n ∈z . или cosx -1/2=0 ⇔cosx =1/2 ⇒ x = ±π/3 +2πk , k ∈ z. ответ : π/2+πn ,n ∈z ; ±π/3 +2πk , k ∈ z. 2 способ cos(x+60°)+ cos(90° -(x+30°) ) =1+cos2x ; cos(x+60°) +cos(60°- x) =1+cos2x ; 2cos60°*cosx =2cos²x ; cosx = 2cos²x ; дальше как в 1 способе * * * * * * * p.s. * * * * * * *cos(α+β) =cosαcosβ - sinαsinβ ; sin(α+β) =sinαcosβ + cosαsinβ ; cos2x =cos²x -sin²x = 2cos²x - 1⇒1+cos2x =2cos²x ; . cos(90° - α) =sinα cosα+cosβ= 2cos(α+β)/2 *cos(α-β)/2 .
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