Сos4x+10sinx/cosx :1/cos²x=3
cos4x+10sinx/cosx *cos²x=3
cos4x+5sin2x=3
1-2sin²2x+5sin2x-3=0
2sin²2x-5sin2x+2=0
sin2x=a
2a²-5a+2=0
D=25-16=9
a1=(5-3)/4=1/2⇒sin2x=1/2⇒2x=(-1)^n*π/6+πn⇒x=(-1)^n*π/12+πn/2
a2=(5+3)/4=2⇒sin2x=2 нет решения
x=π/12∈[-π/6;π/6]