Question

Помогите пожалуйста решить!!!​

Answer 1

$Cos^{2} x+Cos^{2}2x+Cos^{2}3x=1\\\\\dfrac{1+Cos2x}{2}+\dfrac{1+Cos4x}{2} +\dfrac{1+Cos6x}{2}=1\\\\1+Cos2x+1+Cos4x+1+Cos6x=2\\\\Cos2x+Cos4x+Cos6x=-1\\\\2Sinx\cdot Cos2x+2Sinx\cdot Cos4x+2Sinx\cdot Cos6x=-1\cdot 2Sinx\\\\2SinxCos2x+2Sinx Cos4x+2Sinx Cos6x+ 2Sinx=0\\\\-Sinx+Sin3x-Sin3x+Sin5x-Sin5x+Sin7x+2Sinx=0\\\\Sinx+Sin7x=0\\\\2Sin\dfrac{x+7x}{2} Cos\dfrac{x-7x}{2} =0\\\\Sin4x Cos3x=0$

$\left[\begin{array}{ccc}Sin4x=0\\Cos3x=0\end{array}\right\\\\\\\left[\begin{array}{ccc}4x=\pi n,n\in Z \\3x=\dfrac{\pi }{2}+\pi n,n\in Z \end{array}\right\\\\\\\left[\begin{array}{ccc}x=\dfrac{\pi n }{4} ,n\in Z\\x=\dfrac{\pi }{6}+\dfrac{\pi n }{3},n\in Z \end{array}\right$