3.2. sin^2x+0,5sin2x = 1
(1-sqrt(2))^2=1-2sqrt(2)+2=3-2sqrt(2)
то
[m]\sqrt[6]{3-2\sqrt{2}}=\sqrt[6]{(1-\sqrt{2})^{2}}=\sqrt[3]{1-\sqrt{2}}[/m]
[m]\sqrt[3]{1+\sqrt{2}}\cdot \sqrt[6]{3-2\sqrt{2}}=\sqrt[3]{1+\sqrt{2}}\cdot \sqrt[3]{1-\sqrt{2}}= \sqrt[3]{1^{2}-(\sqrt{2})^{2}}=[/m]
[m]= \sqrt[3]{-1}=-1[/m]
О т в е т. -1
sin^2x+0,5*2sinxcosx=sin^2x+cos^2x
sinxcosx=cos^2x
sinxcosx-cos^2x=0
cosx(sinx-cosx)=0
cosx=0 ⇒ x=[m]\frac{\pi}{2}[/m]+2πn, n ∈ Z
или
sinx-cosx=0 ⇒ tgx=1 ⇒ x=[m]\frac{\pi}{4}[/m]+πk, k ∈ Z
О т в е т.[m]\frac{\pi}{4}[/m]+πk, [m]\frac{\pi}{2}[/m]+2πn, k,n ∈ Z