Find mistake in solving $\sin 2x=\sin x + \cos x$ I am solving $\sin 2x = \sin x +\cos x $ for $0\le x \le 360$
$$\sin 2x = \sin x +\cos x$$
$$2 \sin x \cos x =\sin x + \cos x$$
$$(\cos x + \sin x) ^{2} - (\sin x)^{2} - (\cos x)^{2} =(\cos x + \sin x)^{2} - 1=\sin x + \cos x$$
Let $\cos x + \sin x = y$
$$y^{2} - 1= y$$
After solving the quadratic gets $1.618$(I think this is not accepted)  and $-0.618$
$$y=\cos x + \sin x =\sin 2x = -0.618$$
$$2x=218.17,321.83,578.17,681.87$$
$$x=109.09,160.92,289.09,340.34$$
But the problem is $109.09$ and $340.34$ is not the solution. I didn't purposely square anything to produce extra solution. The two extra solution satisfy $\sin 2x=-0.618$ but not $\cos x + \sin x=-0.618$.Is there any mistake in my calculations, or is there anyplace I introduced accidentally extra solution? Thanks. 

Consider these three equations:
\begin{align}
\sin 2x = \sin x + \cos x \tag{1} \\
(\sin x + \cos x)^2 - 1 = \sin x + \cos x \tag{2} \\
(\sin 2x)^2 -1 = \sin 2x \tag{3}
\end{align}
You were correct in determining that (1) and (2) are equivalent. Also, (1) does imply (3).
However, what you failed to notice is that (3) does not imply (1), so while all solutions to 
\begin{align}
y^2 - 1 &= y, \\
y &= \sin x + \cos x
\end{align}
are solutions to (1), the same isn't true for
\begin{align}
y^2 - 1 &= y, \\
y &= \sin 2x.
\end{align}

To illustarte this, here is a example using only basic algebra and not trigonometry, which might look stupid, but is logically analogous to your soliution:
Problem: Solve 
$$
x^2 - x = x - 1. \tag{1}
$$
First, we add $1-x$ to both sides to see that this is equivalent to
$$
(x-1)^2 = 0. \tag{2}
$$
Then, we denote $y=x-1$ to solve $y^2 = 0$ it and get $y=0$. So far so good.
Finally, we use $y = x^2 - x$  (This is a stupid way to do this but it illustrates the error.), and solve
$$
x^2 - x = 0,
$$
and find $x=1$ and $x=0$ as the solutions. 
Where did we go wrong?
