Question

sin2x=1-cos2x
с решение пж!

Answer 1

$sin2x=1-cos2x\ sin2x + cos2x = 1\ frac{1}{ sqrt{2} } sin2x + frac{1}{ sqrt{2} } cos2x = frac{1}{ sqrt{2} } } \ cos frac{ pi}{4} sin2x + sin frac{ pi}{4}cos2x = frac{1}{ sqrt{2} } } \ sin( frac{ pi}{4}+2x)=frac{1}{ sqrt{2} } }\ frac{ pi}{4}+2x=(-1)^karcsinfrac{1}{ sqrt{2} } }+ pi k, k in Z\ frac{ pi}{4}+2x=(-1)^kfrac{ pi}{4}+ pi k, k in Z\ 2x=(-1)^kfrac{ pi}{4}-frac{ pi}{4}+ pi k, k in Z\ x=(-1)^kfrac{ pi}{8}-frac{ pi}{8}+ frac{pi k}{2} , k in Z$