Rezolvare pas cu pas!

[tex]\it {sin2x=2sinxcosx}\\ \rule{80}{0.9} \\ \\ \it (2sinx+cosx)^2=2+3sin^2x \Rightarrow 4sin^2x+4sinxcosx+cos^2x-3sin^2x=2 \Rightarrow \\ \\ \Rightarrow sin^2x+cos^2x+4sinxcosx=2 \Rightarrow 1+4sinxcosx=2\Big|_{-1} \Rightarrow \\ \\ \Rightarrow 4sinxcosx=1\big|_{:2} \Rightarrow 2sinxcosx=\dfrac{1}{2} \Rightarrow sin2x=\dfrac{1}{2}[/tex]