Question

Sin^4x+sin^4(x+пи/4)=1/4 решите пажалуйста

Answer 1

$Sin^{4} x+Sin(x+\frac{\pi }{4})=\frac{1}{4}\\\\(\frac{1-Cos2x}{2})^{2} +(\frac{1-Cos(2x+\frac{\pi }{2}) }{2} )^{2}=\frac{1}{4}\\\\\frac{1}{4}(1-Cos2x)^{2}+\frac{1}{4}(1+Sin2x)^{2}=\frac{1}{4}\\\\1-2Cos2x+Cos^{2}2x+1+2Sin2x+Sin^{2}2x=1\\\\2-2(Cos2x+Sin2x)+\underbrace{(Cos^{2}2x+Sin^{2}2x)}_{1}-1=0\\\\2-2(Cos2x-Sin2x)=0\\\\Cos2x-Sin2x=1\\\\\underbrace{1-Cos2x}_{2Sin^{2}x}+Sin2x=0\\\\2Sin^{2}x+Sin2x=0\\\\2Sin^{2}x+2Sinx Cosx=0\\\\Sinx(Sinx+Cosx)=0$

$\left[\begin{array}{ccc}Sinx=0\\Sinx+Cosx=0|:Cosx,Cosx\neq0 \end{array}\right\\\\\\\left[\begin{array}{ccc}x=\pi n,n\in Z \\tgx+1=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\pi n,n\in Z \\tgx=-1\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\pi n,n\in Z \\x=-\frac{\pi }{4}+\pi n,n\in Z \end{array}\right$