Номер 10 с подробным решением!

$\displaystyle 2 \sqrt{2}Sin^22.5x+Sin3x=Cos3x+ \sqrt{2} 2 \sqrt{2}Sin^22.5x=Cos3x-Sin3x+ \sqrt{2}$

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$sin^2 \frac{x}{2}= \frac{1-cosx}{2}$

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$\displaystyle \sqrt{2}(1-Cos5x)= \sqrt{2}(Cos3x* \frac{ \sqrt{2}}{2}-Sin3x* \frac{ \sqrt{2}}{2}+1)$

$\displaystyle 1-Cos5x=Cos(3x+ \frac{ \pi }{4})+1$

$\displaystyle Cos(3x+ \frac{ \pi }{4})+Cos5x=0$

$\displaystyle 2Cos( \frac{8x+ \pi /4}{2})*Cos( \frac{-2x+ \pi /4}{2})=0$

$\displaystyle \left \{ {{Cos(4x+ \frac{ \pi }{8})=0} \atop {Cos( \frac{ \pi }{8}-x)=0}} \right.$

$Cos(4x+ \pi /8)=0 4x+ \pi /8=+/- \pi /2+2 \pi n, n\in Z 4x=3 \pi /8+2 \pi n, n\in Z; 4x=-5 \pi /8+2 \pi n, x=3 \pi /32+ \pi n/2, n\in Z; x=-5 \pi /32+ \pi n/2, n\in Z$

$Cos( \pi /8-x)=0 \pi /8-x=+/- \pi /2+2 \pi n, n\in Z -x=3 \pi /8+2 \pi n,n\in Z; -x=-5 \pi /8+2 \pi n,n\in Z x=- 3 \pi /8+2 \pi n,n\in Z; x=5 \pi /8+2 \pi n,n\in Z$