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 )