Question

sin^4x-cos^4x=sin2x-1/2

Answer 1

$sin^4x-cos^4x=sin2x-frac{1}{2}\\underbrace {(sin^2x-cos^2x)}_{-cos2x}underbrace {(sin^2x+cos^2x)}_{1}=sin2x-frac{1}{2}\\-cos2x=sin2x-frac{1}{2}\\sin2x+cos2x=frac{1}{2}\\sqrt2cdot (frac{1}{sqrt2}sin2x+frac{1}{sqrt2}cos2x)=frac{1}{2}; ,; ; ; ; frac{1}{sqrt2}=frac{sqrt2}{2} \\sqrt2cdot (cosfrac{pi}{4}cdot sin2x+sinfrac{pi}{4}cdot cos2x)=frac{1}{2}\\sqrt2cdot sin(2x+frac{pi}{4})=frac{1}{2}\\sin(2x+frac{pi}{4})=frac{1}{2sqrt2}$

$2x+frac{pi}{4}=(-1)^{n}cdot arcsinfrac{1}{2sqrt2}+pi n,; nin Z\\2x=-frac{pi}{4}+(-1)^{n}cdot arcsinfrac{1}{2sqrt2}+pi n,; nin Z\\x=-frac{pi}{8}+frac{(-1)^{n}}{2}cdot arcsinfrac{1}{2sqrt2}+frac{pi n}{2},; nin Z$