Extra solutions when solving $\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$

My solution: $$\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$$

$$\sin\theta+\cos\theta=\sqrt{2\times2\sin\theta\cos\theta}$$

$$\sin\theta+\cos\theta=2\sqrt{\sin\theta\cos\theta}$$

$$\sin\theta-2\sqrt{\sin\theta\cos\theta}+\cos\theta=0$$

$$(\sqrt{\sin\theta}-\sqrt{\cos\theta})^2=0\tag{*}$$

$$\sqrt{\sin\theta}=\sqrt{\cos\theta}$$

$$\sqrt{\tan\theta}=1\tag#$$

$$\tan\theta=1$$

$$\theta=n\pi+\frac{\pi}{4}$$

1. You introduced the extraneous solutions in step $(\#),$ because $$\frac{\sqrt{\sin\theta}}{\sqrt{\cos\theta}}=1\implies\sqrt{\frac{\sin\theta}{\cos\theta}}=1;\\\sqrt{\frac{\sin\theta}{\cos\theta}}=1\kern.6em\not\kern -.6em \implies\frac{\sqrt{\sin\theta}}{\sqrt{\cos\theta}}=1.$$ This is because $\frac{\sqrt{\sin\theta}}{\sqrt{\cos\theta}}\not\equiv\sqrt{\tan\theta}.$ By allowing $\sin\theta$ and $\cos\theta$ to be both negative (as well as both positive), you expanded the set of candidate solutions.

   Nevertheless, $(\#)$ is a valid step, as the forward implication is correct.

2. However, the above may be moot, since the prior step $(*)$ may have discarded solutions and thus be invalid: without justification, it is not apparent that $$\sin\theta-2\sqrt{\sin\theta\cos\theta}+\cos\theta=0\implies(\sqrt{\sin\theta}-\sqrt{\cos\theta})^2=0,$$ since $\sin\theta-2\sqrt{\sin\theta\cos\theta}+\cos\theta\not\equiv\sin\theta-2\sqrt{\sin\theta}\sqrt{\cos\theta}+\cos\theta\equiv(\sqrt{\sin\theta}-\sqrt{\cos\theta})^2.$ (The converse is certainly true though.)

   Here, the forward-implication does turn out to be justifiable: $\theta$ happens to reside in the first quadrant, so $\sin\theta$ and $\cos\theta$ are indeed both positive.

Addendum at OP's request: $$\sin\theta+\cos\theta=\sqrt{2\sin2\theta}\\ \implies 1+\sin2\theta=2\sin2\theta\\ \iff\sin2\theta=1$$ $$\iff\theta=(4n+1)\frac\pi4.\tag1$$ But there's the implicit condition that $$\sin\theta+\cos\theta\geq0\\ \sqrt2\sin\left(\theta+\frac\pi4\right)\geq0\\ \theta+\frac\pi4\in\bigcup[2n\pi,(2n+1)\pi]$$ $$\theta\in\bigcup[(8n-1)\frac\pi4,(8n+3)\frac\pi4]\tag2.$$ Combining $(1)$ and $(2)$: $$\theta=(8k+1)\frac\pi4\\=\frac\pi4+2k\pi.$$

$\sin \theta+\cos \theta$ is given to be positive square root of $2 \sin (2 \theta)$. Odd values of $n$ make $\sin \theta+\cos \theta$ negative, so only even values are permitted.

We need that $\sin2\theta\ge 0$ that is

$$0+2n\pi\le2\theta \le \pi+2n\pi \quad \iff \quad n\pi\le \theta \le \frac \pi 2+n\pi$$

therefore only solutions in the first or third quadrant are allowed but since in the third quadrant $\sin \theta + \cos \theta <0$, only solutions in the $\color{red}{\text{first quadrant}}$ are allowed that is by $n=2k$

$$\color{red}{2k\pi\le \theta \le \frac \pi 2+2k\pi}$$

Then, since all terms involved in the expression are positive, and we have that for $A,B\ge 0$

$$A=\sqrt B \iff A^2=B$$

we can then proceed by squaring both side to obtain an $\color{magenta}{\text{equivalent equation}}$

$$\sin\theta+\cos\theta=\sqrt{2\sin2\theta} \iff 1+\sin 2\theta =2 \sin 2\theta \iff \color{magenta}{\sin 2\theta =1} $$

$$\iff2\theta=\frac \pi 2+2n\pi \iff \theta=\frac \pi 4+n\pi\iff \color{magenta}{\theta=\frac \pi 4+2k\pi}$$

Edit

As an alternative approach, since all terms involved in the expression are positive, by AM-GM we obtain

$$\sin\theta+\cos\theta \ge 2\sqrt{\sin \theta \cos \theta}=\sqrt{2\sin 2\theta}$$

with equality for $\sin\theta=\cos\theta \implies \theta=\frac \pi 4+2k\pi$.

Since $\theta$ must lie in the 1st quadrant in order to make both $\sqrt{\cos \theta} $and $\sqrt{\sin \theta}$ real, therefore the general solution of the equation should be $\theta= 2 n\pi+\frac{\pi}{4} \text {, where } n \in \mathbb{Z}.$