Первое вот, а второе не знаю
По формуле квадрат разности получим 0.01х^5-0.09y^2
(^ - значит встепени)
2 - 3cos4x -sin2x = 0 ; x ∈[-π/8 ; 5π<span>/8 ] .
</span>* * * cos2α=cos²α -sin²α = <span> 1-sin²α -sin</span>²α =1 -2sin²<span>α * * * </span>
* * * cos4x =cos2*(2x) = 1 -2sin²2x <span> * * *</span>
2 - 3(1-2sin²2x) -sin2x = <span>0 ;
</span>6sin²2x -sin2x -1 = 0 ;
6t² -t -1<span> = </span>0 <span> ;</span> * * * D =1² -4*6*(-1) =25 =5² <span> * * *
</span>t₁= (1-5)/(2*6) = -1/3<span> ;</span>
t₂= (1+5)/12 = 1/2.
а) sin2x=1/2 ;
[ 2x = π/6 +2πn ; 2x =(π -π/6) +2πn , n∈Z.
[ x = π/12 +πn ; x =5π/12+πn , n∈Z.
учитывая условия x ∈ [-π/8;5π/8 ] , получается <span>[x = π/12 </span>; x=5π/12.
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б) sin2x=<span> -1/3</span>⇔2x =(-1)^(n+1) arcsin(1/3)+πn, n∈Z.
[ x = -(1/2)arcsin(1/3) + πn ; x=(1/2)*(-π+arcsin(1/3)+πn, n∈Z;
ответ: -(1/2)arcsin(1/3) ; π/6 ; 5π/12 <span>.
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