Question

решите плз!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

$frac{cos^22x-sin^2x}{1-sin3x}=0\ {1-sin3x}neq0Rightarrowsin3xneq1Rightarrow3xneqpi nRightarrow xneqfracpi3n,;;ninmathbb{Z}\cos^22x-sin^2x=0\cos2x=1-2sin^2xRightarrowcos^22x=\=(1-2sin^2x)^2=1-4sin^2x+4sin^4x\1-4sin^2x+4sin^4x-sin^2x=0\4sin^4x-5sin^2x+1=0\sin^2x=t,;;sin^4x=t^2,;;tin[0,;1]\4t^2-5t+1=0\D=25-16=9\t_1=1,;;t_2=frac14\sin^2x=1Rightarrowcos x=pm1\x_1=fracpi2+2pi n,;;x_2=frac{3pi}2+2pi n,;;ninmathbb{Z}$
$sin^2x=frac14Rightarrowsin x=pmfrac12\x_3=fracpi6+2pi n,;;x_4=frac{7pi}6+2pi n,;;ninmathbb{Z}$