Корень из 3*cos2x+sin2x=1

Метод вспомогательного аргумента

√3cos2x+sin2x=1 |:2

√3/2cos2x+1/2sin2x=1/2

sin (π/3) cos2x+cos (π/3) sin2x=1/2

sin (π/3+2x) = 1/2

π/3+2x=π/6+2πn, n∈Z |-π/3 π/3+2x=5π/6+2πn, n∈Z |-π/3

2x=-π/6+2πn, n∈Z |:2 2x=3π/6+2πn, n ∈Z |:2

x=-π/12+πn/, n ∈Z x=π/4+πn, n∈Z