How could I sieve through the extra solutions to $\sin\theta +\cos\theta=\sin2\theta $in the interval $[-\pi,\pi]$? 
I am trying to find the number of solutions of the equation $$\sin\theta +\cos\theta=\sin2\theta $$ in the interval $[-\pi,\pi]$.

Here's what I did, $$\sin\theta +\cos\theta=\sin2\theta \\ \Rightarrow (\sin\theta +\cos\theta)^2=(\sin2\theta)^2 \\ \Rightarrow1+\sin2\theta=(\sin2\theta)^2$$
Hence we get $\sin2\theta=\frac{1\pm\sqrt{5}}{2}$, with $\sin2\theta=\frac{1 -\sqrt{5}}{2}$ bein the only valid solution.
Since $2\theta$ is present inside $\sin()$, I assumed that there would be four solutions of the equation. But on plotting the graph I find that there are only two solutions.
Why are there only two solution instead of four, and how could I prevent this mistake in future?
 A: By only considering $\sin 2\theta=\dfrac{1-\sqrt{5}}{2}$, one gets four roots since $\sin 2\theta$ has a period of $\pi$ and the interval $[-\pi,\pi]$ is twice that length.
But one also has to consider the equality of $\sin \theta + \cos \theta =\dfrac{1-\sqrt{5}}{2}$. Since
$$\sin \theta + \cos \theta = \sqrt{2}\sin \left( \frac{\pi}{4} + \theta \right)$$
has a period of $2\pi$, actual number of roots is only $2$.
A: OP: how can I figure out the correct number of solutions without plotting a graph?
How can I prevent or atleast sieve though the extra solutions? The solutions of this equation itself are too complicated to check manually.

*

*Short of plugging the candidate solutions into the original
equation, there is no general method for directly sifting out
extraneous ones or determining how many there are.
(If a calculator is allowed, it is typically a good-enough check to
just plug in the approximated candidate solutions.)


*However, it is possible to catch potential extraneous solutions:
carefully watch every line for any step that is not obviously
“reversible”,
noting that every extraneous solution arises from such a step
(which is not to say that every such step creates an extraneous
solution):

*

*neglecting domain restrictions typically of trigonometric and logarithmic functions


*squaring an equation whose two sides don't identically have the same sign


*converting absolute value to $\pm$
$\lvert x+6\rvert=2x\\\pm(x+6)=2x\\x=-2 \;\text{ or }\; 6 \text{    (the negative solution is extraneous)}$


*inadvertently multiplying an equation by $0$
$x=1\\x^2=x\\x=0 \;\text{ or }\; 1$


*performing some other non-injective operation on an equation.
Method 1 above is easier and safer than Method 2.
A: If you use the multiple angle formula $\theta=2 \tan^{-1}(x)$
$$\sin(\theta )+\cos(\theta)-\sin(2\theta)=0 \implies x^4-6 x^3+2 x-1=0$$ which has only two real solutions
$$x_1=\frac{3+\sqrt{5}}{2}+\sqrt{\frac{1}{2} \left(11+5 \sqrt{5}\right)}$$
$$x_2=\frac{3+\sqrt{5}}{2}-\sqrt{\frac{1}{2} \left(11+5 \sqrt{5}\right)}$$
Then $\theta_1 \sim 2.80847$ and $\theta_2 \sim -1.23768$ that you observed plotting.
Be always careful when you square.
A: Avoid squaring whenever practicable as it often introduces
.  When do we get extraneous roots?
. Extraneous Roots
et  $\sin\theta+\cos\theta=t,t^2=?\le1+1$
$t=t^2-1, 2t=\pm\sqrt5+1$
Now  clearly $(\sqrt5+1)^2>2$
$\implies\cos\left(\theta-\dfrac\pi4\right)=\dfrac t{\sqrt2}=?$
As $\cos(2\pi-y)=\cos y$ and cosine has a period $=2\pi,$ we shall have two incongruent solutions
A: The squaring generated extra solutions. To see this , consider this simple example : $x=3$ has trivially a unique solution, but $x^2=9$ has $-3$ and $3$.
