[tex]\it x\in \Big(0,\ \dfrac{\pi}{2}\Big),\ \ sinx=\dfrac{3}{4}\ \Rightarrow cosx=\sqrt{1-\Big(\dfrac{3}{4}\Big)^2}=\sqrt{1-\dfrac{9}{16}}=\sqrt{\dfrac{7}{16}}=\dfrac{\sqrt7}{4}\\ \\ \\ sin2x+cos2x=2 sinx cosx+1-2sin^2x =2\cdot\dfrac{3}{4}\cdot\dfrac{\sqrt7}{4}+1-2\cdot\dfrac{9}{16}=\\ \\ \\ =\dfrac{3\sqrt7}{8}+1-\dfrac{9}{8}=\dfrac{3\sqrt7}{8}-\dfrac{1}{8}=\dfrac{3\sqrt7-1}{8}[/tex]
