Question

Решить: cos^2(6x)+sin^2(3x)-1=0

Answer 1

$cos^2(6x)+sin^2(3x)-1=0\\boxed{sin^2alpha=frac{1-cos2alpha}2}\\2cos^2(6x)+1-cos(6x)-2=0\2cos^2(6x)-cos(6x)-1=0\D=1+8=9\cos6x=1 || cos6x=-frac{1}2\6x=2pi k || 6x=pmfrac{2pi}3+2pi k\underline{x=frac{pi k}3} || underline{x=pmfrac{pi}9+frac{pi k}3}\kin Z$