3) 2sin²(3π/2+x) = √3cosx ; x∈[ -7π/2 ; -2π].
2*(-cosx)² = √3cosx ;
2cos²x - √3cosx = 0;
2cosx(cosx - √3/2) = 0 ;
[ cosx =0 ; cosx =√3/2.
[ x =π/2 + π*k ; x = - π/6+2π*k ; x = π/6+2π*k ; k∈Z.
Выбор корней x∈[ -7π/2 ; -2π] :
- 7π/2 ≤ π/2 + π*k ≤ -2π ⇔ -7π/2 - π/2 ≤ π*k ≤ -π/2 -2π ⇔ -4π ≤ π*k ≤ -5π/2 ⇔
- 4≤k≤ -5/2 ⇒k = - 4 ; -3⇒x = -7π/2 ; x = -5π/2.
из x = - π/6+2π*k ⇒x = - 13π/6 при k = - 1.
из x = π/6+2π*k нет решения x∈[ -7π/2 ; -2π] .
ответ : -7π/2 ; -5π/2 ; -13π/6 .
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24) 2sin²x - 3sinx +1 =0 ; x∈ (3π ; 9π/2] .
[ sinx =1/2 ; sinx =1. *** sinx = t ***
[ x = ( -1)^(k)*π/6 +π*k ; x =π/2+ 2π*k , k ∈ Z .
Для выбора корней удобно писать решения в виде :
[ x = π/6 +2π*k ; x =(π -π/6) +2π*k ; x =π/2+ 2π*k , k ∈ Z.
или тоже самое :
[ x = π/6 +2π*k ; x =5π/6) +2π*k ; x =π/2+ 2π*k , k ∈ Z.
ответ : π/6 +4π = 25π/4 ; x =π/2+ 4π =9π/2.
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45) cosx(2cosx + tqx) =1 ; x∈ [ -5π/2 ; -π/2] .
2cos²x + sinx =1 ;
2(1 -sin²x) +sinx = 1;
2sin²x - sinx -1 =0 ;
[sinx = -1/2 ; sinx =1 ;
[ x = (-1)^(k+1) π/6 +π*k ; x =π/2 +2π*k ;k ∈ Z.
Для выбора корней удобно писать решения в виде :
[ x = - π/6 + 2π*k ; x = -5π/6 +2π*k ; x =π/2 +2π*k ;k ∈ Z.
[ x = - π/6 - 2π ; x= π/2 - 2π = - 3π/2 .
ответ : - π/6 -2π = - 13π/6 ; π/2- 2π = - 3π/2.