Решите систему: {x-y=2 {x^2-xy+y^2=7

2sinxcosx + sinx - 2cosx -1 = 0 2cosx (sinx - 1) +sinx -1 = 0 (2cosx +1)(sinx - 1) = 0 2cosx + 1 = 0                         sinx - 1 = 0 cosx = -1/2                               sinx = 1 x  ∈(2π/3 + 2πn; n∈z)            x  ∈(π/2 +2πk; k∈z)  sin3x = -cosx  sin²3x = cos²x (1-cos6x) /2 =(1+cos2x) /2     1 - cos6x - 1 - cos2x = 0 cos6x + cos2x = 0 2cos[(6x+2x)/2] cos[ (6x-2x)/2] = 0 2cos4xcos2x = 0  cos4x = 0                               cos2x  = 0  4x =  π/2+2πn; n∈z               x =  π/4 +  πn; n∈z x =  π/8 +  π/2n; n∈z

1-2sina=2(1/2-sina)=2(sin30· -·sina)=4sin(15·-a/2)cos(15·+a/2) 1+sina=1+cos(90 ·-a)=2cos^2(45·-·a/2)    или по другому 1+sina=sin90· +sina=2sin(45·+a/2)cos(45·-a/2)=2cos(90·-(45·+a/2))cos(45·-a/2)= =2cos(45·-a/2)*cos(45·-a/2)=2cos^2(45·-·a/2) доказать тождество: cosa +cos (2п/3 +a) + cos( 2п/3 - a) = 0 cosa +cos (2п/3 +a) + cos( 2п/3 - a) = cosa +2cos((2п/3 +a) + ( 2п/3 - a))/2* *cos((2п/3 +a) -( 2п/3 - a))/2=cosa +2cos2п/3*cosa= =cosa +2cos(п-п/3)*cosa=cosa -2cosп/3*cosa=cosa -2*1/2*cosa=cosa -cosa=0