Question

1-sin3x=(sin x|2-cos x|2)в квадрате

Answer 1

$1-sin 3x=(sin frac{x}{2} -cosfrac{x}{2} )^2\ \ 1-sin3x=sin^2frac{x}{2} +cos^2frac{x}{2} -2sinfrac{x}{2} cos frac{x}{2} \ \ 1-sin3x=1-sin x\ \ sin3x-sin x=0\ \ 2sin frac{3x-x}{2} cos frac{3x+x}{2} =0\ \ 2sin xcos 2x=0\ \ left[begin{array}{ccc}sin x=0\\ cos 2x=0end{array}rightRightarrow left[begin{array}{ccc}x_1= pi k,k in mathbb{Z}\ \ x_2= frac{pi}{4}+ frac{pi n}{2},n in mathbb{Z} end{array}right$