Question

2 вариант, кто сколько сможет

Answer 1

$1);2sin^2x+5cos x+1=0\2-2cos^2x+5cos x+1=0\2cos^2x-5cos x-3=0\cos x=t,;cos^2x=t^2,;-1leq tleq1\2t^2-5t-3=0\D=25+4cdot2cdot3=49\t_1=-frac12\t_2=3;-;He;nogx\cos x=-frac12Rightarrow x=frac{2pi}3+2pi n,;ninmathbb{Z}$

$2);sin3xcos x=sin xcos3x\frac{sin2x+sin4x}2=frac{sin(-2x)+sin4x}2\sin2x+sin4x=-sin2x+sin4x\2sin2x=0\sin2x=0\2x=pi n\x=fracpi2n,;ninmathbb{Z}$

$3);cos2x-cos x=0\2cos^2x-1-cos x=0\2cos^2x-cos x-1=0\cos x=t,;cos^2x=t^2,;-1leq tleq1\2t^2-t-1=0\D=1+4cdot2=9\t_1=-frac12,;t_2=1\cos x=-frac12Rightarrow x=frac{2pi}3+2pi n\cos x=1Rightarrow x=pi n,;ninmathbb{Z}$

$4);cos(0,5pi+2x)+sin x=0\cosleft(fracpi2+2xright)+sin x=0\-cos2x+sin x=0\-1+2sin^2x+sin x=0\2sin^2x+sin x-1=0\sin x=t,;sin^2x=t^2,;-1leq tleq1\2t^2+t-1=0\D=1+4cdot2=9\t_1=-1,;t_2=frac12\sin x=-1Rightarrow x=(-1)^ncdotfrac{3pi}2+pi n\sin x=frac12Rightarrow x=(-1)^ncdotfracpi6+pi n,;ninmathbb{Z}$

$5);sin^4x+cos2x=1\sin^4x+1-2sin^2x=1\sin^4x-2sin^2x=0\sin^2x=t,;sin^4x=t^2,;0leq tleq1\t^2-2t=0\t(t-2)=0\t_1=0,\t_2=2;-;HE;nogx.\sin^2x=0Rightarrowsin x=0Rightarrow x=pi n,;ninmathbb{Z}$