Question

1-cos4x=sin2x помогите пожалуста!

Answer 1

$1-cos 4x=sin2x\ 1-(1-2sin^22x)=sin2x\ 1-1+2sin^22x=sin2x\ 2sin^22x-sin2x=0\ sin2x(2sin2x-1)=0\ \ left[begin{array}{ccc}sin2x=0\ sin2x=0.5end{array}rightRightarrow left[begin{array}{ccc}x_1= frac{pi k}{2},k in Z\ x_2=(-1)^kcdot frac{pi}{12} + frac{pi k}{2},k in Z end{array}right$