Question

Алебра -10 ,№1256 (4)
cos²x+cos²2x= sin²3x+sin²4x

Answer 1

$Cos^{2}x+Cos^{2} 2x=Sin^{2}3x+Sin^{2}4x\\\\\dfrac{1+Cos2x}{2} +\dfrac{1+Cos4x}{2} =\dfrac{1-Cos6x}{2} +\dfrac{1-Cos8x}{2}\\\\1+Cos2x+1+Cos4x=1-Cos6x+1-Cos8x\\\\(Cos2x+Cos8x)+(Cos4x+Cos6x)=0\\\\2Cos\dfrac{2x+8x}{2}Cos\dfrac{2x-8x}{2} +2Cos\dfrac{4x+6x}{2}Cos\dfrac{4x-6x}{2}=0 \\\\Cos5xCos3x+Cos5xCosx=0\\\\Cos5x(Cos3x+Cosx)=0$

$Cos5x\cdot2Cos\dfrac{3x+x}{2}Cos\dfrac{3x-x}{2}=0\\\\Cos5x Cos2x Cosx=0\\\\\left[\begin{array}{ccc}Cos5x=0\\Cos2x=0\\Cosx=0\end{array}\right \\\\\\\left[\begin{array}{ccc}5x=\dfrac{\pi }{2}+\pi n,n\in Z \\2x=\dfrac{\pi }{2}+\pi n,n\in Z \\x=\dfrac{\pi }{2}+\pi n,n\in Z \end{array}\right \\\\\\\left[\begin{array}{ccc}x=\dfrac{\pi }{10}+\dfrac{\pi n }{5} ,n\in Z \\x=\dfrac{\pi }{4}+\dfrac{\pi n }{2} ,n\in Z \\x=\dfrac{\pi }{2}+\pi n,n\in Z \end{array}\right$