[tex]\it 4)\ \ x\in\Big(\dfrac{\pi}{2},\ \pi\Big) \Rightarrow cosx<0\ \ \ \ \ \ (*)\\ \\ \\ sin^2x+cos^2x=1 \Rightarrow cos^2x=1-sin^2x\stackrel{(*)}{\Longrightarrow}\ cosx=-\sqrt{1-\Big(\dfrac{3}{4}\Big)^2} \Rightarrow\\ \\ \\ cosx=-\sqrt{\dfrac{7}{16}}=-\dfrac{\sqrt7}{4}[/tex]
[tex]\it sinx+cosx =\dfrac{3}{4}+\dfrac{\sqrt7}{4}=\dfrac{3+\sqrt7}{4}[/tex]
