Вопрос школьника по предмету Алгебра
решить уравнение Sin3x+sinx/sinx-1=0
Ответ учителя по предмету Алгебра
sin3x +sinx/(sinx -1) =0 ;
ОДЗ : sinx ≠ 1 ⇔ x ≠π/2 +πn , n∈Z
sinx(3 -4sin²x) +
sinx/(sinx -1) =0 ;
sinx*( 3 — 4sin²x +1/(sinx -1) ) =0 ;
a) sinx =0 ⇒ x =πn ; n∈Z.
b) 3 — 4sin²x +1
/(sinx -1) =0 ;
4sin³x — 4sin²x -3sinx +2 =0 ; * * *sinx =1/2 * * *
4sin³x — 2sin²x —
2sin²x +sinx —4sinx +2 =0
;
2sin²x(2sinx-1) -sinx(2sinx-1) -2(2sinx -1) =0 ;
(2sinx-1)(2sin²x -sinx -2) =0 ⇔[2sinx 1=0 ; 2sin²x -sinx -2 =0;
b₁)
2sinx — 1=0 ;
sinx =1/2 ⇒ x= (-1)^n *π/6+πn , n∈Z.
b₂)
2sin²x -sinx -2 =0 ; замена: t =sinx , -1 ≤ t ≤1
2t² — t -2 =0 D =1² -4*2(-2) =(√17)² ;
t₁ = (1 + √17)/2*2 =(1 +
√17)/4 > 1_не удовлетворяет
t₂ = (1 — √17) /4⇒
sinx = — (√17 -1)/4 ⇒
x =( -1)^(n+1)*arcsin (1 — √17) /4 +πn , n∈Z.
ответ :
x =πn;
(-1)^(n) *π/6+πn ; ( -1)^(n+1)*arcsin (1 — √17) /4 +πn , n∈Z.
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