Question

sin6x+sin2x+2sin^2x=1

Answer 1

$sin6x+sin2x+2sin^2x=1,\ 2sinfrac{6x+2x}{2}cosfrac{6x-2x}{2}-1+2sin^2x=0,|*(-1), backslash {sina+sinb=2sinfrac{a+b}{2}cosfrac{a-b}{2}}\ -2sin4xcos2x+1-2sin^2x=0, backslash {cos2a=cos^2a-sin^2a=1-2sin^2a=2cos^2a-1 }$

$\ cos2x-2sin4xcos2x=0,\ cos2x(1-2sin4x)=0,\ left [ {{cos2x=0} atop {sin4x=1/2}} right.; Rightarrow left [ {{2x=frac{pi }{2}+pi k} atop {4x=(-1)^nfrac{pi }{6}+pi n}} right.; Rightarrow left [ {{x=frac{pi}{4}+frac{pi k}{2} } atop {x=(-1)^nfrac{pi}{24}+frac{pi n}{4} }} right.;n,k in Z$