2 cos^2x +14 cos x=3 sin^2x

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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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