Finding when $y=1+2\sin(x)$ and $y=2\sin(\frac{x}{2})+2\cos(\frac{x}{2})$ are equal How would you solve the following equation?
$$1 +2\sin(x) = 2\sin(\tfrac{x}{2}) +2\cos(\tfrac{x}{2}) \quad\text{for }−2\pi\leq x \leq 2\pi.$$
Steps I tried:
\begin{align}
1+2\sin(x) &= 2\sin(\tfrac{x}{2}) +2\cos(\tfrac{x}{2}) \\
2\sin(x) -2\sin(\tfrac{x}{2}) &= 2\cos(\tfrac{x}{2}) -1 \\
2[\sin(x) -\sin(\tfrac{x}{2})] &= 2\cos(\tfrac{x}{2}) -1 \\
\sin(x) -\sin(\tfrac{x}{2}) &= \cos(\tfrac{x}{2}) -\tfrac{1}{2} \\
\sin(x) +\tfrac{1}{2} &= \cos(\tfrac{x}{2}) +\sin(\tfrac{x}{2})
\end{align}
Now, squaring both sides, 
$$\sin^2(x) +\tfrac{1}{4} = \cos^2(\tfrac{x}{2}) +\sin^2(\tfrac{x}{2}).$$
Using $\sin^2(x) +\cos^2(x) = 1$, 
\begin{align}
\sin^2(x) &= 1 -\tfrac{1}{4} \\
\sin(x) &= \mp\sqrt\frac{3}{4}.
\end{align}
When I follow through, I get the solutions below the $x$-axis but not above. Can anyone please explain what's wrong with my solution?

Examining your attempt
Your first five lines do nothing but divide the original equation by $2$.

$$1+2\sin(x)=2\sin(\frac{x}{2})+2\cos(\frac{x}{2})$$

$2\sin(x)-2\sin(\frac{x}{2})=2\cos(\frac{x}{2})-1$
$2[\sin(x)-\sin(\frac{x}{2})]=2\cos(\frac{x}{2})-1$
$\sin(x)-\sin(\frac{x}{2})=\cos(\frac{x}{2})-\frac{1}{2}$

$$\sin(x)+\frac{1}{2}=\cos(\frac{x}{2})+\sin(\frac{x}{2})$$

You then make a classic mistake when squaring. This line is incorrect.

$$\sin^2(x)+\frac{1}{4}=\cos^2(\frac{x}{2})+\sin^2(\frac{x}{2})$$

When squaring both sides of an equation you should square the entire left hand side and the entire right hand side not just each individual bit. If you could square each bit then we could write $1+1=2\implies1^2+1^2=2^2$ which is clearly not true. Instead if you squared your step would look like:
$$\left(\sin(x)+\frac{1}{2}\right)^2=\left(\cos(\frac{x}{2})+\sin(\frac{x}{2})\right)^2$$
Solution
If we expand the brackets above we get:
$$\sin^2(x)+\sin(x)+\frac14=\cos^2(\frac{x}{2})+2\cos(\frac{x}{2})\sin(\frac{x}{2})+\sin^2(\frac{x}{2})$$
$$\sin^2(x)+\sin(x)+\frac14=1+\sin(x)$$
$$\sin^2(x)=\frac34$$
This is the same as what you had but it is a coincidence. Your working is still incorrect (or at least very very brief and skipping some important working out).
$$\sin(x)=\pm\frac{\sqrt{3}}{2}$$
$$x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\text{ for }0\leq x\leq2\pi$$
We need to substitute these back in and check each answer as taking the square root could introduce extraneous answers of which we find one. Hence we get:
$$x=\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{3}\text{ for }0\leq x\leq2\pi$$
Obviously there exists other solutions which are these plus integer multiple of $2\pi$.
