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FeniksHeat
@FeniksHeat
July 2022
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решите 4 и 5 номера пожалуйста завтра контрольная:(
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oganesbagoyan
Verified answer
4.
а)
2cos²x -3cosx +1 =0 ;
Квадратное уравнение относительно cosx || можно и замену t=
cosx ||
[ cosx =1 ; cosx =1/2⇔ [
x =2πn ; x =±π/3 + 2πn , n∈Z
===========
б)
4sin²x +4cosx -1 =0 ;
4(1-cos²x) +
4cosx -1 =0 ;
4cos²x - 4cosx -3 =0 ; * * * cos²x - cosx -3/4 =0 * * *
cosx =( 2± √(2² -4*(-3) ) /4 =( 2±
4) /4
[ cosx = 3 /2 ; cosx = -1/2 ;
cosx = 3
/2 > 1 _не имеет решения
cosx = -1/2 ⇒
x = ±2π/3 + 2πn , n∈Z .
5.
√3sin4x +cos4x =0 ;
2(√3 /2 *sin4x +1/2*cos4x) =0 ;
2(√cos(π/6) *sin4x +sin(π/6)
*cos4x) =0 ;
2sin(4x+π/6) =0 ⇔sin(4x+π/6) =0⇒4x+π/6 =πn , n
∈Z ;
x = -
π/24 + (π/4)*n , n
∈Z .
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Answers & Comments
Verified answer
4.а) 2cos²x -3cosx +1 =0 ;
Квадратное уравнение относительно cosx || можно и замену t=cosx ||
[ cosx =1 ; cosx =1/2⇔ [ x =2πn ; x =±π/3 + 2πn , n∈Z
===========
б) 4sin²x +4cosx -1 =0 ;
4(1-cos²x) + 4cosx -1 =0 ;
4cos²x - 4cosx -3 =0 ; * * * cos²x - cosx -3/4 =0 * * *
cosx =( 2± √(2² -4*(-3) ) /4 =( 2± 4) /4
[ cosx = 3 /2 ; cosx = -1/2 ;
cosx = 3 /2 > 1 _не имеет решения
cosx = -1/2 ⇒ x = ±2π/3 + 2πn , n∈Z .
5.
√3sin4x +cos4x =0 ;
2(√3 /2 *sin4x +1/2*cos4x) =0 ;
2(√cos(π/6) *sin4x +sin(π/6)*cos4x) =0 ;
2sin(4x+π/6) =0 ⇔sin(4x+π/6) =0⇒4x+π/6 =πn , n∈Z ;
x = - π/24 + (π/4)*n , n∈Z .