Question

sin2xsin4x=cos2x

найти x​

Answer 1

$Sin2x Sin4x=Cos2x\\\\\frac{1}{2}[Cos(2x-4x)-Cos(2x+4x)]-Cos2x=0\\\\Cos2x-Cos6x-2Cos2x=0\\\\Cos6x +Cos2x=0\\\\2Cos\frac{6x+2x}{2}Cos\frac{6x-2x}{2}=0\\\\Cos4xCos2x=0\\\\\left[\begin{array}{ccc}Cos4x=0\\Cos2x=0\end{array}\right$

$\left[\begin{array}{ccc}4x=\frac{\pi }{2}+\pi n,n\in Z \\2x=\frac{\pi }{2}+\pi n,n\in Z \end{array}\right\\\\\\\boxed{\left[\begin{array}{ccc}x=\frac{\pi }{8}+\frac{\pi n }{4} ,n\in Z \\x=\frac{\pi }{4}+\frac{\pi n }{2} ,n\in Z \end{array}\right}$