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