3^sin2x*27^cosx=1 [0;5]

$3^{sin2x} * 27^{cosx} =1 3^{2sinx*cosx} * (3^{3} ) ^{cosx} =1 3^{2six*cosx} * 3^{3cosx} =1 3^{2sinxcosx+3cosx} = 3^{0}$
2sinxcosx+3cosx=0
cosx*(2sinx+3)=0
cosx=0 или 2sinx+3=0
x=π/2+2πn, n∈Z 2sin=-3. решений нет

0≤π/2+πn≤5 | (-π/2)
-π/2≤πn≤5-π/2 |: π
-1/2≤n≤5/π-1/2
-1/2≤n≤1
n=0;1
x₁=π/2+π*2, x₁=π/2
x₂=π/2+π*1, x₂=3π/2