Question

Решите уравнение: $sin^{4}x +sin^{4}(x+\pi /4)= \frac{1}{4}$

Answer 1

$Sin^{4} x+Sin^{4}\Big(x+\dfrac{\pi }{4}\Big)= \dfrac{1}{4}\\\\\Big(\dfrac{1-Cos2x}{2} \Big)^{2} +\Big(\dfrac{1-Cos(2x+\frac{\pi }{2}) }{2}\Big)^{2}=\dfrac{1}{4}\\\\\dfrac{(1-Cos2x)^{2} }{4}+\dfrac{(1+Sin2x)^{2} }{4}=\dfrac{1}{4} \ |\cdot \ 4\\\\(1-Cos2x)^{2} +(1+Sin2x)^{2}=1$

$1-2Cos2x+Cos^{2}2x+1+2Sin2x+Sin^{2}2x=1\\\\\underbrace{Sin^{2}2x+Cos^{2}2x}_{1}+2(Sin2x-Cos2x)=1-2\\\\2(Sin2x-Cos2x)=-2\\\\Sin2x-Cos2x=-1 \ | \ :\sqrt{2} \\\\\dfrac{1}{\sqrt{2} }Sin2x-\dfrac{1}{\sqrt{2} }Cos2x=-\dfrac{1}{\sqrt{2} }\\\\Cos\dfrac{\pi }{4}Sin2x-Sin\dfrac{\pi }{4}Cos2x=-\dfrac{1}{\sqrt{2} } \\\\Sin\Big(2x-\dfrac{\pi }{4 }\Big)=-\dfrac{1}{\sqrt{2} }$

$2x-\dfrac{\pi }{4}=(-1)^{n} arcSin\Big(-\dfrac{1}{\sqrt{2} }\Big)+\pi n,n\in Z\\\\2x-\dfrac{\pi }{4}=(-1)^{n} \Big(-\dfrac{\pi }{4} \Big)+\pi n,n\in Z\\\\2x-\dfrac{\pi }{4}=(-1)^{n+1}\dfrac{\pi }{4}+\pi n,n\in Z\\\\2x=(-1)^{n+1}\dfrac{\pi }{4}+\dfrac{\pi }{4}+\pi n,n\in Z\\\\x=(-1)^{n+1}\dfrac{\pi }{8}+\dfrac{\pi }{8}+\dfrac{\pi n }{2} ,n\in Z$