Способом разложения ** множители: 4sin⁴(x-π)+cos²(3π/2 -2x)=2 [-5π/2; -3π/2]

Question

Способом разложения на множители:
4sin⁴(x-π)+cos²(3π/2 -2x)=2 [-5π/2; -3π/2]

Answer 1

4Sin⁴(x-π) + Cos²(3π/2-2x) = 2
4Sin⁴x + Sin²2x = 2
4Sin⁴x + 4Sin²xCos²x = 2
4Sin²x(Sin²x + Cos²x) = 2
4Sin²x = 2
2Sin²x = 1
Sin²x = 1/2
Sinx = √2/2                                                 Sinx = - √2/2
x = (-1)ⁿarcSin√2/2 + πn, n ∈ z                  x = (-1)ⁿarcSin(-√2/2) + πn, n ∈ z
x = (-1)ⁿπ/4 + πn, n ∈ z                              x = (-1)ⁿ(-π/4) + πn, n ∈ z
                                                                   x = (-1)ⁿ⁺¹π/4 + πn, n ∈ z
Корни из заданного промежутка : - 7π/4 ,  - 9π/4