Помогите пожалуйста хоть что-нибудь! Заранее благодарен.

1(1-cos2x)²/4+(1+cos2x)²/4=5/82-4cos2x+2cos²2x+2+4cos2x+2cos²2x-5=04cos²2x-1=0(2cos2x-1)(2cos2x+1)=0cos2x=1/2⇒2x=+-π/3+2πn⇒x=+-π/6+πn,n∈zcos2x=-1/2⇒2x=+-2π/3+2πk⇒x=+-π/3+πk,k∈z2(cos3x-cos7x)+(sin2x-sin6x)=02sin5x*sin2x-2sin2x*cos4x=02sin2x*(sin5x-cos4x)=0sin2x=0⇒2x=πn⇒x=πn/2,n∈zsin5x-sin(π/2-4x)=02sin(9x/2-π/4)*cos(x/2+π/4)=0sin(9x/2-π/4)=0⇒9x/2-π/4=πk⇒9x/2=π/4+πk⇒x=π/18+2πk/9,k∈zcos(x/2+π/4=0⇒x/2+π/4=π/2+πt⇒x/2=π/4+πt⇒x=π/2+2πt,t∈z31/2cos(x/2-3x/2)+1/2cos(x/2+3x/2)-1/2cos(x-3x)+1/2cos(x+3x)-1/2cos(2x-3x)+1/2cos(2x+3x)=01/2cosx+1/2cos2x-1/2cos2x+1/2cos4x-1/2cosx+1/2cos5x=01/2(cos4x+cos5x)=01/2*2cos(9x/2)*cos(x/2)=0cos(9x/2)=0⇒9x/2=π/2+πn⇒x=π/9+2πn/9,n∈zcos(x/2)=π/2+πk⇒x=π+2πk,k∈z45sin²x-4sinxcosx-cos²x-4sin²x-4cos²x=0/cos²xtg²x-4tgx-5=0tgx=aa²-4a-5=0a1+a2=4 U a1*a2=-5a1=-1⇒tgx=-1⇒x=-π/4+πn,n∈za2=5⇒tgx=5⇒x=arctg5+πk,k∈z