
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Let's note that $\sin x \neq 0$ and $\cos x \neq 0$, and multiply both parts of the equation by $\operatorname{tg} x$. We get $\sqrt{2+\cos 2 x+\sqrt{3} \operatorname{tg} x}=\sin x+\sqrt{3} \cos x$. Given that $\sin x+\sqrt{3} \cos x \geq 0$, both sides of this equation can be squared. Since $\sin x+\sqrt{3} \cos x=2 \sin \left(x+\frac{\pi}{3}\right)$, the inequality holds if $-\frac{\pi}{3}+2 \pi n \leq x \leq \frac{2 \pi}{3}+2 \pi n, n \in \mathbb{Z}$. Under these constraints and conditions $\sin x \neq 0, \cos x \neq 0$, the equation is equivalent to:

\[ 
2+\cos 2 x+\sqrt{3} \operatorname{tg} x=\sin ^{2} x+2 \sqrt{3} \sin x \cos x+3 \cos ^{2} x, \operatorname{tg} x-2 \sin x \cos x=0, \frac{\sin x}{\cos x}-2 \sin x \cos x=0, \sin x\left(\frac{1}{\cos x}-2 \cos x\right)=0 
\]

Thus, we arrive at the equation: 
\[ \cos ^{2} x=\frac{1}{2}, \cos x= \pm \frac{\sqrt{2}}{2}, x= \pm \frac{\pi}{4}+\pi n, n \in \mathbb{Z} \]

Taking into account the constraints, we obtain the solutions to the original equation: 
\[ x= \pm \frac{\pi}{4}+2 \pi n, \quad n \in \mathbb{Z} \]

Present the answer in LaTex format: \boxed{Your answer}
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