1) cosx=cos3x 2) sinx+cos3x=0

1)сosx-cos3x=0cosa-cosb=-2sin(a-b)/2*sin(a+b)/22sinxsin2x=0sinx=0⇒x=πnsin2x=0⇒2x=πn⇒x=πn/2Ответ x=πn/2,n∈Z2)sinx+sin(π/2-3x)=0sina+sinb=2sin(a+b)/2*cos(a-b)/2=02sin(π/4-x)cos(π/4+2x)=0sin(π/4-x)=cos(x+π/4)=0⇒x+π/4=π/2+πn⇒x=π/4+πncos(π/4+2x)=0⇒2x+π/4=π/2+πn⇒2x=π/4+πn⇒x=π/8+πn/2

[latex]cos x=cos 3x\ cos x=4cos^3x-3cos x\ 4cos x-4cos^3x=0\ 4cos x(1-cos ^2x)=0\ cos x=0\ x= frac{pi}{2}+ pi n,n in Z\ 1-cos^2x=0\ sin^2x=0\ x= pi k,k in Z [/latex][latex]sin x+cos3x=0\ | sqrt{1-cos^2x}|=4cos^3x-3cos x [/latex] Пусть [latex]cos x=t,(|t| leq 1)[/latex][latex]| sqrt{1-t^2}|=4t^3-3t [/latex] ОДЗ: 1-t²≥0[latex]1-t^2=(4t^3-3t)^2[/latex] Пусть t² = z (z≥0)[latex]1-z-16z^3+24z^2-9z=0\ 16z^3-24z^2+10z-1=0\ 16z^3-8z^2-16z^2+8z+2z-1=0\ 8z^2(2z-1)-8z(2z-1)+(2z-1)=0\ (2z-1)(8z^2-8z+1)=0\ z_1=0.5\ \ 8z^2-8z+1=0\ D=b^2-4ac=64-32=32\ z_2_,_3= frac{2pm sqrt{2} }{4} [/latex]Возвращаемся от z[latex]t^2=0.5\ t=pm frac{ sqrt{2} }{2} [/latex][latex]t^2= frac{2pm sqrt{2} }{4} \ t=pm frac{ sqrt{2pm sqrt{2} } }{2} [/latex][latex]t=frac{ sqrt{2} }{2} [/latex] не удовлетворяет ОДЗВозвращаемся к замене[latex]cos x=-frac{ sqrt{2} }{2} \ x=pm frac{3 pi }{4}+2 pi n,n in Z\ \ cos x=pm frac{ sqrt{2+ sqrt{2} } }{2} \x=pmarccos(pm frac{ sqrt{2+ sqrt{2} } }{2} )+2 pi n,n in Z[/latex]