С1 по cos2x+sin^2x=0,26. решить уравнение ,найти корни [3п; 9п/2]

Cos2x+sin²x=0,26

сos²x - sin²x + sin²x = 0,26

сos²x = 0,26

1) cosx₁ = √0.26                               2) cosx₁ = -√0.26

x₁₁ = arccos (√0.26 ) + 2πn                    x₂₁ = π - arccos (√0.26 ) + 2πn

x₁₂= -arccos (√0.26 ) + 2πn                    x₂₂ = -π + arccos (√0.26 ) + 2πn

находим корни в промежутке х∈[3π; 4,5π]

п = 1  x₁₁ = arccos (√0.26 ) + 2π < 3π            х₁₁∉[3π; 4,5π]

          x₁₂= -arccos (√0.26 ) + 2π < 2π           x₁₂∉[3π; 4,5π]

          x₂₁ = π - arccos (√0.26 ) + 2π < 3π      x₂₁∉[3π; 4,5π]

          x₂₂ = -π + arccos (√0.26 ) + 2π < 2π     x₂₂∉[3π; 4,5π]

п = 2  x₁₁ = arccos (√0.26 ) + 4π                    х₁₁∈[3π; 4,5π]

          x₁₂= -arccos (√0.26 ) + 4π                   x₁₂∈[3π; 4,5π]

          x₂₁ = π - arccos (√0.26 ) + 4π > 4,5π   x₂₁∉[3π; 4,5π]

          x₂₂ = -π + arccos (√0.26 ) + 4π            x₂₂∈[3π; 4,5π]

п = 3  x₁₁ = arccos (√0.26 ) + 6π > 6π            х₁₁∉[3π; 4,5π]

          x₁₂= -arccos (√0.26 ) + 6π > 5π           x₁₂∉[3π; 4,5π]

          x₂₁ = π - arccos (√0.26 ) + 6π > 6π      x₂₁∉[3π; 4,5π]

          x₂₂ = -π + arccos (√0.26 ) + 6π > 5π     x₂₂∉[3π; 4,5π]

ответ: x₁₁ = arccos (√0.26 ) + 4π                  

           x₁₂= -arccos (√0.26 ) + 4π                 

            x₂₂ = 3π + arccos (√0.26 )