Question

РЕшить уравнение (81^sinx)^cosx=(1/9)^корень из2*cosx

Answer 1

$(81^{sin x})^{cos x}= (frac{1}{9} )^{sqrt{2}cosx } \ \ 9^{2sinxcosx}=9^{- sqrt{2}cosx }$
2sin x cos x =-√2 cos x
2 sin x cos x + √2 cos x = 0
cos x (2 sin x + √2) = 0
1) cos x = 0
    x = π/2 + πn, n∈Z
2) 2 sin x + √2 = 0
    sin x = -√2/2
    x = -π/4 + 2πk, k∈Z

Ответ: x₁ = π/2 + πn; x₂ = -π/4 + 2πk;  n,k∈Z