Question

решите уравнение 2 cos^2x-sin4x=1

Answer 1

2cos²x-sin4x=1
(2cos²x-1)-sin4x=0
cos2x-2*sin2x*cos2x=0
cos2x*(1-2sin2x)=0
cos2x=0 или 1-2sin2x=0
1. cos2x=0, 2x=π/2+πn, n∈Z |:2
x₁=π/4+πn/2, n∈Z

2. 2sin2x=1, sin2x=1/2
$2x=(-1) ^{n}*arcsin frac{1}{2}+ pi n,$n∈Z
$2x=(-1) n^{2} * frac{ pi }{6} + pi n,$n∈Z|:2

$x_{2} =(-1) ^{2} * frac{ pi }{12} + frac{ pi n}{2} , n$∈Z