Помогите решить: 1) 2sin6x*cos2x=sin8x+1 2) 4sinxcos(pi/2+5x)=1-cos4x 3) sin7x+cos^2 2x=sin^2 2x+sinx

1) 2 * sin 6x * cos2x = sin 8x + 1 ;2 * 1 / 2 (sin ( 6 x + 2 x) + sin ( 6 x - 2 x ) ) = sin 8x + 1 ;sin 8x + sin 4x = sin 8x + 1 ;sin 8x + sin4x - sin8x - 1 = 0 ;sin 4x - 1 = 0 ;sin 4x = 1;4 * x = pi / 2 + 2 * pi * n , n ∈ Z ;x=pi / 2 / 4 + 2 * pi * n / 4 , n ∈ Z ; x = pi / 8 + pi * n / 2, n ∈ Z.2) 4 * sin x * cos ( pi / 2 + 5 * x ) = 1 - cos 4x ;- 4 * sinx * sin5x = 2 - 2 cos ² 2x ;- 4/2 * (cos ( x - 5 x ) -cos ( x+ 5x) ) = 1 - cos4x ;2 * cos6x - 2cos4x + cos4x = 1 ;2* cos 6x - cos 4x = 1 ;2 * cosx = cos4x+1 ;2 * cos4x * cos2x-2sin4x * sin2x=2cos^ 2 2x ;2 * cos2x * (2cos^ 2 ( 2x-1 ))-4 * cos2x * sin^ 2x- 2 * cos^ 2 2x=0;4 * cos^ 3 2x-2cos2x-4cos2x(1-cos^ 2 2x)-2cos^ 2 2x=0;8 * cos ^ 3 2x-2 * cos^ 2 2x-6 * cos2x=0;2 * cos2x * (4cos^ 2 2x-cos2x-3) = 0;cos2x=0;2x=pi /2+pi * n;x=pi /4+pi n/2,n ∈ Z;4cos^ 2 2x-cosx-3=0;Пусть cos2x=a, тогда;4 * a ^ 2 - a - 3 = 0;a1 = - 3 / 4, тогда ;2x= + - ( pi - arccos 3/4 ) + 2 * pi / n;x = + - 1 / 2 * (pi -arccos3/4)+pi * n;a2=1, тогда;cos2x=2 * pi * n;x= pi * n,n∈z;3) sin7x+cos^2 2x=sin^2 2x+sinx ; sin7x-sinx+cos^ 2 2x-sin^ 2 2x=0 ;2 * sin3x * cos4x + cos4x = 0 ;cos4x * ( 2 * sin3x + 1) = 0 ;cos4x=0 ; 4 * x= pi / 2 + pi * n ;x=pi /8+pi * n / 4, n ∈ z ;sin3x=-1/2 ;3 * x=(-1)^ n+1*pi/6+pi * n ;x=(-1)^n+1*pi/18+pi * n,n∈z.