Question

решить ............................................

Answer 1

Ответ:

$sin^2\Big(\dfrac{x}{4}+\dfrac{\pi}{4}\Big)\cdot sin^2\Big(\dfrac{x}{4}-\dfrac{\pi}{4}\Big)=0,375\cdot sin^2\Big(-\dfrac{\pi}{4}\Big)\\\\\\\Big(\, sin\Big(\dfrac{x}{4}+\dfrac{\pi}{4}\Big)\cdot sin\Big(\dfrac{x}{4}-\dfrac{\pi}{4}\Big)\, \Big)^2=\dfrac{3}{8}\cdot \Big(-sin\dfrac{\pi}{4}\Big)^2\\\\\\\dfrac{1}{4}\cdot \Big(cos\dfrac{\pi}{2}-cos\dfrac{x}{2}\Big)^2=\dfrac{3}{8}\cdot \Big(-\dfrac{\sqrt2}{2}\Big)^2\\\\\\\Big(0-cos\dfrac{x}{2}\Big)^2=\dfrac{3\cdot 4}{8}\cdot \dfrac{2}{4}$

$cos^2\dfrac{x}{2}=\dfrac{3}{4}\ \ ,\ \ \ \ \dfrac{1+cosx}{2}=\dfrac{3}{4}\ \ ,\ \ \ 1+cosx=\dfrac{3}{2}\ \ ,\ \ cosx=\dfrac{1}{2}\ ,\\\\\\Otvet:\ \ x=\pm \dfrac{\pi}{3}+2\pi n\ ,\ n\in Z\ .\\\\\\\star \ \ sin\alpha \cdot \sin\beta =\dfrac{1}{2}\cdot \Big(cos(\alpha -\beta)-cos(\alpha +\beta)\Big)\ \ \star \\\\\\\star \ \ cos^2\alpha =\dfrac{1+cos2\alpha }{2}\ \ \star$