task/29461152
2cos²x +14cosx = 3sin²x ⇔2cos²x +14cosx = 3(1 -cos²x) ⇔5cos²x +14cosx - 3 =0 ⇔ [ cosx = -3 < -1 ; cosx= 1/5. ⇒ cosx= 1/5 ⇔ x =± arccos(1/5)+2πn , n ∈ ℤ .
ответ : ± arccos(1/5)+2πn , n ∈ ℤ .
* * * 5cos²x +14cosx - 3 =0⇔ [ 5cos²x +15cosx - cosx - 3 =0 ⇔
5cosx(cosx +3) - (cosx + 3 )=0 ⇔(cosx +3)(5cosx -1 )=0 ⇔ 5(cosx +3)(cosx -1)=0. или D₁= 7² -5*(-3) =8² ; cosx = (-7 ±8) /5 . * * *
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