How to simplify this surd: $\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}$ $$\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}} = x$$
We have to find the value of $x$. 
Taking the terms to other side and squaring is increasing the power of $x$ rapidly, and it becomes unsolvable mess.
I think the answer lies in simplification, but can't do it. Also I have tried taking $\sqrt{2}$ common, but it doesn't help.
 A: Hint: square the surd and note the cross term may be simplified.
Also note that $\sqrt{3}/2 = \cos{(\pi/6)}$ and you may use the double angle formula $1 +\cos{t} = 2 \cos^2{(t/2)}$, etc.
A: Recognize the quantity under radical sign is a perfect square, it all cancels out.
$$\sqrt{1+\frac{\sqrt{3}}{2}}  = \sqrt{1+2\cdot \frac12 \cdot \frac{\sqrt{3}}{2}} = \sqrt { (\cos \pi/3 + \sin \pi/3 )^2} = (\cos \pi/3 + \sin \pi/3 ) $$
Similarly, 
$$\sqrt{1-\frac{\sqrt{3}}{2}}  = (\cos \pi/3 - \sin \pi/3 ) $$
Adding, $$ \rightarrow \sqrt 3 $$
A: try converting expr inside surd into a square.
$$x = \sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}$$
$$ = \sqrt{\frac{4+2\sqrt{3}}{4}}+\sqrt{\frac{4-2\sqrt{3}}{4}} $$
$$ = \sqrt{\frac{1+3+2\sqrt{3}}{2^2}}+\sqrt{\frac{1+3-2\sqrt{3}}{2^2 }} $$
$$ = \sqrt{\left(\frac{1+\sqrt{3}}{2}\right)^2}+\sqrt{\left(\frac{1-\sqrt{3}}{2}\right)^2} $$
$$ = \frac{|1+\sqrt{3}|}{2} + \frac{|1-\sqrt{3}|}{2}$$ 
$$ = \frac{1+\sqrt{3}}{2} + \frac{\sqrt{3}-1}{2}$$ 
$$ = \sqrt{3}$$
Edit
My earlier answer was incorrect since i foolishly took square root and forgot about modulus. I have ammended it. It is important to note that: 
$$\sqrt{x^2} = |x|$$
A: Squaring the equation:
\begin{equation}
x^2=2+2\sqrt{1-\frac{3}{4}}=2+1=2+1=3
\end{equation}
Finally you get
$$x=\sqrt{3}$$
