Step 1:
Concept:
Using half angle theorem
[tex]\begin{gathered} \text{cosx = cos(}\frac{x}{2}+\frac{x}{2}\text{)} \\ =cos^2(\frac{x}{2})-sin^2(\frac{x}{2}) \\ \sin ^2(\frac{x}{2})+cos^2(\frac{x}{2})\text{ = 1} \\ \sin ^2(\frac{x}{2})=1-cos^2(\frac{x}{2}) \\ \text{Therefore} \\ =cos^2(\frac{x}{2})-1+cos^2(\frac{x}{2}) \\ =2cos^2(\frac{x}{2})\text{ - 1} \end{gathered}[/tex]
Hence
[tex]\text{cosx = 2cos}^2(\frac{x}{2})\text{ - 1}[/tex]
Step 2:
[tex]\begin{gathered} \sqrt[]{2\cos x\text{ + 2}} \\ =\text{ }\sqrt[]{2(2\cos ^2(\frac{x}{2})\text{ - 1) + 2}} \\ =\text{ }\sqrt[]{4cos^2(\frac{x}{2})\text{ -2 + 2}} \\ =\text{ }\sqrt[]{4\cos ^2(\frac{x}{2})} \\ =\text{ 2cos}\frac{x}{2} \end{gathered}[/tex]
Final answer
[tex]2\cos \frac{x}{2}[/tex]
