1+ 2 sin4x+3cos^2 2x=0 (, как можно быстрее)

1 + 2 sin4x+3cos^2(2x)=0

cos^2(2x) + sin^2(2x) + 4sin2xcos2x +  3cos^2(2x) = 0

sin^2(2x)+ 4sin2xcos2x + 4cos^2(2x) = 0   |: cos^2x

tg^2(2x) + 4tg2x + 4 = 0

(tg2x + 2)^2 = 0

tg2x + 2 = 0

tg2x = -2

2x = arctg(-2) + пk

x = -0,5arctg2 + 0,5пk

1)   cos5α-cosα = -2sin2α*sin3α ; 2) cosα - cos3α +2sin2α   = 2sinα  *sin2α +2sin2α =2sin2α*(sinα+1) = 2sin2α(1 +cos(π/2 -α))   =4sin2α*cos² (π/4   -  α/2).   cosα - cos3α +2sin2α   = 2sinα  *sin2α +2sin2α =2sin2α*(sinα+1) = 2sin2α(sinα+sinπ/2) =4sin2α*sin(π/4   +α/2)*cos(π/4 -α/2)   =   [[   но   ***   sin(π/4   +α/2) =cos(π/2 -(π/4   +α/2)) =cos(π/4 -  α/2)   *** ]] = 4sin2α*cos² (π/4   -  α/2).

(cos 7x   -cos3x)  +(sin8x+sin2x) =0; -  2sin2xsin5x +2sin5xcos3x   =  0; sin5x(sin2x  -cos3x) =0 ; sin5x=0  ⇒ x =π*k/5 ,k  ∈z . sin2x  -cos3x =0 sin2x -sin(π/2 -3x) =0; 2sin(5x/2 -π/4)*cos(π/4 -x/2) =0 ; sin(5x/2 -π/4)*cos(x/2 -π/4 ) =0 ; sin(5x/2 -π/4) =0⇒5x/2   -π/4 =π*k     ⇔          x  =  π/10+2π/5*k,k  ∈z  cos(x/2 -π/4)=  0   ⇒x/2  -  π/4   =π/2 +π*k  ⇔     x=3π/2+2π*k ,k∈z *********************************************************************** sin2x -cos3x=0; 2sinxcosx -cosx(4cos²x -3) =0; ² cosx(2sinx -4cos²x +3) =0 cosx(4sin²x +2sinx   -1) =0