Question

cos x sin 9x = cos 3x sin 7x помогите с подробным решением​

Answer 1

$CosxSin9x=Cos3xSin7x\\\\\frac{1}{2}[(Sin(9x-x)+Sin(9x+x)]=\frac{1}{2}[Sin(7x-3x)+Sin(7x+3x)]\\\\Sin8x+Sin10x=Sin4x+Sin10x\\\\Sin8x+Sin10x-Sin4x-Sin10x=0\\\\Sin8x-Sin4x=0\\\\2Sin4xCos4x-Sin4x=0\\\\Sin4x(2Cos4x-1)=0$

$\left[\begin{array}{ccc}Sin4x=0\\2Cos4x-1=0\end{array}\right\\\\\left[\begin{array}{ccc}4x=\pi n,n\in Z \\Cos4x=\frac{1}{2} \end{array}\right\\\\\left[\begin{array}{ccc}x=\frac{\pi n }{4},n\in Z \\4x=\pm arcCos\frac{1}{2}+2\pi n,n\in Z \end{array}\right\\\\\left[\begin{array}{ccc}x=\frac{\pi n }{4},n\in Z \\4x=\pm \frac{\pi }{3}+2\pi n,n\in Z  \end{array}\right\\\\\left[\begin{array}{ccc}x=\frac{\pi n }{4},n\in Z \\x=\pm \frac{\pi }{12}+\frac{\pi n }{2},n\in Z \end{array}\right$