Task/25765851
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Решить уравнение
(1/9) ^ cos(π/2 -x) =3^(2sin(x+π/2) ; [ -7π/2 ; -2π]
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a)
(1/9) ^( cos(π/2 -x) ) =3^(2sin(π/2+x) ;
(1/9) ^ (sinx) =3^(2cosx) ;
(3⁻²) ^ (sinx) =3^(2cosx) ;
(3) ^ (-2sinx) =3^(2cosx) ;
-2sinx) =2cosx ; || : -cosx ≠0
tgx = - 1;
x = -π/4 +πn , n∈ℤ. (общее решение)
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b) x∈ [ -7π/2 ; -2π]
-7π/2 ≤ -π/4 +πn≤ - 2π ;
-7π/2 + π/4 ≤ πn≤ π/4 - 2π ;
-13π/4 ≤ πn ≤ -7π/4 ;
-13/4 ≤ n ≤ -7/4 , т.е. n= - 3 ; - 2. (два решения на отрезке [ -7π/2 ; -2π] )
x₁ = - π/4 - 3π = -13π/4 ;
x₂ = - π/4 - 2π = - 9π/4.
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Удачи !