решите уравнение
а)[tex]cos3x= \sqrt{2} /2
[/tex]
б)[tex]3cos ^{2} x+cosx-4=0
[/tex]
в)[tex] \sqrt{3}cos2x+sin2x=0
[/tex]

2)решите неравенство
[tex]sinx> \sqrt{2}/2 [/tex]

a) cos3x=√2/23x= (+∨-)π/4+2πKX=(+∨-)π/12+2/3*πKб)  3cos²x+cosx-4=0 3t² +t -4=0t₁=(-1-sqrt(1-4*3*(-4))/(2*3) =(-1-7)/6= -4/3t₂=(-1+sqrt(1-4*3*(-4))/(2*3) =(-1+7)/6= 1cosx = -4/3 <-1  cosx =1  ==>x=2π*k   ;  k∈Z (любое целое число)в) √3cos2x+sin2x=02(√3/2cos2x + 1/2sin2x)=02(cosπ/6*cos2x + sinπ/6*sin2x)=02cos(2x -π/6) =02x -π/6=π/2 +π*k2x=2π/3+π*kx=π/3+π/3*k ;    k∈Z (любое целое число)2)  sinx >√2/2π/4<x< π-π/4    π/4<x< 3/4π  2π*k+π/4<x< 3/4π +2π*k x∈ (2π*k+π/4x ; 3/4π +2π*k )