Решение
<span>(cos²x - sin²x)• </span>√(<span>1 - х²) = 0
1) cos</span>²x - sin²x = 0
cos2x = 0
2x = π/2 + πk, k ∈ Z
x = π/4 + πk/2, k ∈ Z
2) √(1 - x²) = 0
<span> [√(1 - x²)[</span>²<span> = 0
</span>1 - x² = 0
x² = 1
x₁ = - 1
x₂ = 1
х^2 - 25 - x^2 - x + 3 x - 3 = 9 x
3a-9b-5a+10b=-2a+b
-2*(-1.5)-1=3-1=2
=(y²-9)(y²+9)-(4y^4-4y³+y²)-19=y^4-81-4y^4+4y³-y²-19=-3y^4+4y³-y²-100
Sinx*cosx - 6sinx +6cosx +6 = 0 ;
6(cosx - sinx) + (1 - (cosx - sinx)²)/2 +6 = 0;
*** (cosx -sinx)² =cos²x -2sinx*cosx +cos²x=1-2sinx*cosx ⇒sinx*cosx=(1 - (cosx - sinx)²)/2***
Замена : t = cosx - sinx ;
6t +(1 -t²)/2 +6 =0 ;
t² -12t -13 =0;
t ₁= - 1 ;
t ₂=13 ;
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cosx -sinx = - 1
1/√2*cosx - 1/√2 *sinx = -1/√2 ;
cos(x +π/4) = 1/√2 ;
x+π/4 = (+/-)3π/4 +2π*k ;
x = (+/-)3π/4 -π/4 +2π*k ;
или разделяя :
x₁ = -3π/4 - π/4 +2π*k = - π +2π*k ;
x₂ = 3π/4 - π/4 +2π*k =π/2 + 2π*k , k∈Z.
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сosx - sinx = 13 ; уговаривать x бесполезно !
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ответ : <u> С </u><u>праздником Великой Победы </u><u>!</u> - π +2π*k , π/2 + 2π*k , k∈Z .