√3cosx + sinx = 2cos2x
Разделим обе части данного уравнения на 2
(√3/2)•сosx + (1/2)•sinx = cos2x
sin(π/3)•cosx + cos(π/3)•sinx = cos2x
sinα•cosβ + cosα•sinβ = sin(α + β)
sin( π/3 + x ) = cos2x
sin( π/3 + x ) - cos2x = 0
sin( π/3 + x ) - sin( π/2 - 2x ) = 0
sinα - sinβ = 2•sin( (1/2)•(α - β) )•cos( (1/2)•(α + β) )
2•sin( (1/2)•(π/3 + x - π/2 + 2x) )•cos( (1/2)•(π/3 + x + π/2 - 2x) ) = 0
2•sin( (1/2)•(3x - π/6) )•cos( (1/2)•(-x + 5π/6) ) = 0
Произведение равно нулю, если хотя бы один из множителей равен нулю.
1) sin( (1/2)•(3x - π/6) ) = 0
(1/2)•(3x - π/6) = πn
3x - π/6 = 2πn
3x = π/6 + 2πn
x = π/18 + 2πn/3 , n ∈ Z
2) cos( (1/2)•(-x + 5π/6) ) = 0
(1/2)•(-x + 5π/6) = π/2 + πk
- x + 5π/6 = π + 2πk
x = - π/6 + 2πk , k ∈ Z
ОТВЕТ: π/18 + 2πn/3 , n ∈ Z ; - π/6 + 2πk , k ∈ Z