Question

sin ²3x + sin²4x =sin²5x+ sin²6х
помогите пожалуйста​

Answer 1

Ответ:

$sin^23x+sin^24x=sin^25x+sin^26x\ \ ,\ \ \ \ \ \boxed{\ sin^2a=\dfrac{1-cos2a}{2}\ }\\\\\\\dfrac{1-cos6x}{2}+\dfrac{1-cos8x}{2}=\dfrac{1-cos10x}{2}+\dfrac{1-cos12x}{2}\ \Bid|\cdot 2\\\\\\1-cos6x+1-cos8x=1-cos10x+1-cos12x\ |\cdot (-1)\\\\cos6x+cos8x=cos10x+cos12x\\\\2\cdot cos7x\cdot cosx=2\cdot cos11x\cdot cosx\\\\cos11x\cdot cosx-cos7x\cdot cosx=0\\\\cosx\cdot (cos11x-cos7x)=0\\\\a)\ \ cosx=0\ \ ,\ \ x=\dfrac{\pi}{2}+\pi n\ ,\ n\in Z\\\\b)\ \ cos11x-cos7x=0\ \ ,\ \ \ -2\cdot sin9x\cdot sin2x=0$

$sin9x=0\ \ ,\ \ 9x=\pi k\ \ ,\ \ x_2=\dfrac{\pi k}{9}\ ,\ k\in Z\\\\sin2x=0\ \ ,\ \ 2x=\pi m\ \ ,\ \ x_3=\dfrac{\pi m}{2}\ ,\ m\in Z\\\\x_2=x_3\ ,\ esli\ \ \dfrac{\pi k}{9}=\dfrac{\pi m}{2}\ \ \Rightarrow \ \ k=\dfrac{9m}{2}\ \ \to \ \ m=2l\ \ ,\ \ k=9l\ ,\ l\in Z\\\\Otvet:\ \ x_1=\dfrac{\pi}{2}+\pi n\ \ ,\ \ x_2=\dfrac{\pi k}{9}\ \ ,\ \ n,k\in Z\ .$