Решить уравнения :
cos2x-5cosx-2=0
1-cos8x=sin 4x
sin^2x+4sinx*cosx+3cos^2x=0
cos4x-sin4x=-1/2
xsin^2x+sin^2 3x+sin^2 4x+sin^2 5x=2

1)cos2x-5cosx-2=0;⇒2cos²x-1-5cosx-2=0;⇒cosx=y;-1≤y≤1;⇒2y²-5y-3=0;y₁,₂=(5⁺₋√(25+24))/4=(5⁺₋7)/4;y₁=(5+7)/4=3;⇒y₁>1⇒нет решения;y₂=(5-7)/4=-1/2;⇒cosx=-1/2;⇒x=⁺₋2π/3+2kπ;k∈Z.2)1-cos8x=sin4x;⇒sin²4x+cos²4x-cos²4x+sin²4x=sin4x;⇒2sin²4x-sin4x=0;⇒sin4x(2sin4x-1)=0;⇒sin4x=0⇒4x=nπ;k∈Z;⇒x=nπ/4;n∈Z;2sin4x-1=0;⇒sin4x=1/2;4x=(-1)ⁿ·π/6+nπ;n∈Z.3)sin²x+4sinxcosx+3cos²x=0;⇒cos²x≠0 делим на cos²x:tg²x+4tgx+3=0;tgx=y;⇒y²+4y+3=0;y₁,₂=-2⁺₋√(4-3)=-2⁺₋1;y₁=-1;⇒tgx=-1;⇒x=-π/4+nπ;n∈Z;y₂=-3;⇒x=arctg(-3)+nπ;n∈Z.4)cos4x-sin4x=-1/2;⇒cos4x=sin4x-1/2;⇒cos²4x=sin²4x-2/2·sin4x+1/4;⇒1-sin²4x-sin²4x+sin4x-1/4=0⇒-2sin²4x+sin4x+3/4=0;⇒sin4x=y;-1≤y≤1;2y²-y-3/4=0;y₁,₂=(1⁺₋√(1+6))/4=(1⁺₋√7)/4;y₁=(1+√7)/4=(1+2.646)/4=0.9115;sin4x=0.9115;⇒4x=(-1)ⁿarcsin(0.9115)+2nπ;n∈Z;x=(-1)ⁿ(arcsin(0.9115))/4+nπ/2;n∈Z;y₂=(1-2.646)/4=-0.4115;4x=(-1)ⁿarcsin(-0.4115)+2nπ;n∈Zx=[(-1)ⁿarcsin(-0.4115)+2nπ]/4.