Question

1-sin3x=(sin x|2-cos x|2)в квадрате

Answer 1

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