Решите уравнение пожалуйста а)sin2x+2sinx=√3cosx+√3 б)2cos^2(3П/2+x)=sin2x в) (sin2x+cosx)(√3+√3*√tgx)=0 г)10*5^2x-1-19*35^x+1470*7^2x-2=0
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A)sin2x +2sinx =√3cosx+√3 ;
2sinx*cosx +2sinx =√3cosx+√3 ;
2sinx(cosx +1) - √3(cosx +1) =0 ;
2(cosx +1) (sinx -√3 /2 ) =0 ;
[ cosx = -1 ; sinx = √3 /2⇔ [ x = (1+2n)π ; x = (-1)^n*π/3 + πn ; n∈Z .
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б)
2 cos²(3π/2+x) =sin2x ; * * * cos(3π/2+x) =sinx * * *
2sin²x =sin2x ;
2sinx(sinx -cosx) =0 ;
[ sinx =0 ; sinx =cosx⇔ [ x =πn ; tqx =1 .⇔ [ x =πn ; x =π/4 + πn ; n∈Z .
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в)
(sin2x+cosx)(√3+√3*√tgx)=0 ;
2cosx(sinx+1/2)*√3(1 + √tgx ) =0 ; ОДЗ :tgx ≥0
2√3 *cosx(sinx+1/2)*(1 + √tgx ) =0 ;
1 + √tgx ≥ 1 ≠ 0
cosx ≠0 не определена tgx ;
{ sinx= -1/2 ; tgx ≥0 . ⇒ x =4π/3+ 2πn , ; n∈Z .
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г)
10*5^(2x-1)-19*35^x+1470*7^(2x-2)=0 ;
10(5^x)²* 5⁻¹ - 19*5^x*7^x + 1470*(7^x)²* 7⁻² =0 ;
2*(5^x)² - 19*5^x*7^x + 30*(7^x)² =0 ;
2*( (5/7)^x)² -19*(5/7)^x +30 =0 ; замена : t = (5/7)^x >0
2t² - 19t +30 =0 ;
t₁ =(19-11) /4 = 2 ⇔(5/7) ^x =2 ⇒x₁ =Lq2 /(Lq5 -Lq7).
t ₂ =(19+11) /4 = 15/2 ⇔(5/7) ^x =15/2 ⇒x₂ =(Lq15-Lq2) / (Lq5 -Lq7).