you will receive a question-answer pair and be asked to rewrite just  the answer and explains your reasoning step by step, incorporating dynamic Chain of Thought (CoT), reflection, and verbal reinforcement learning. Follow these instructions:

1. Enclose all thoughts within <thinking> tags, exploring multiple angles and approaches.
2. Break down the solution into clear steps, providing a title and content for each step.
3. After each step, decide if you need another step or if you're ready to give the final answer.
4. Continuously adjust your reasoning based on intermediate results and reflections, adapting your strategy as you progress.
5. Regularly evaluate your progress, being critical and honest about your reasoning process.
6. Assign a confidence score between 0.0 and 1.0 to guide your approach:
   - 0.8+: Continue current approach
   - 0.5-0.7: Consider minor adjustments
   - Below 0.5: Seriously consider backtracking and trying a different approach
7. If unsure or if your score is low, backtrack and try a different approach, explaining your decision.
8. For mathematical problems, show all work explicitly using LaTeX for formal notation and provide detailed proofs.
9. Explore multiple solutions individually if possible, comparing approaches in your reflections.
10. Use your thoughts as a scratchpad, writing out all calculations and reasoning explicitly.
11. Use at least 5 methods to derive the answer and consider alternative viewpoints.
12. Be aware of your limitations as an AI and what you can and cannot do.

### Example Problem:
*Problem*: Solve for \( x \) in the equation \( 2x + 3 = 11 \).

### Expected Format:
*Step 1*: Subtract 3 from both sides of the equation.  
<thinking>Subtracting 3 from both sides should isolate the term with \( x \). Am I confident this is the correct first step? Yes, because it simplifies the equation to \( 2x = 8 \).</thinking>  
*Step 1 Confidence Score*: 0.85

*Step 2*: Divide both sides by 2.  
<thinking>Dividing by 2 will solve for \( x \). This is a straightforward division. Is there any reason to doubt this step? No, it should give \( x = 4 \).</thinking>  
*Step 2 Confidence Score*: 0.90

*Step 3*: Verify the solution by substituting \( x = 4 \) back into the original equation.  
<thinking>Substituting \( x = 4 \) back into the equation should confirm the solution. Does \( 2(4) + 3 = 11 \)? Yes, it does, so the solution is verified.</thinking>  
*Step 3 Confidence Score*: 1.00

Question: Let's consider a function of the form:
$$g(x)=\arcsin x +\arcsin \sqrt{1-x^2}$$
Find the range of $g(x)$, expressed in radians.
Answer: The domain of $\arcsin x$ is $[-1, 1]$, so the domain of $g(x)$ is also $[-1, 1]$.

The function $g(x)$ involves two inverse sine functions, so the range of each component is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Let's analyze the behavior of $g(x)$ at the endpoints of its domain:

1. When $x = -1$, we have
\[ g(-1) = \arcsin(-1) + \arcsin \sqrt{1-(-1)^2} = \arcsin(-1) + \arcsin(0) = -\frac{\pi}{2} + 0 = -\frac{\pi}{2} \]

2. When $x = 1$, we have
\[ g(1) = \arcsin(1) + \arcsin \sqrt{1-(1)^2} = \arcsin(1) + \arcsin(0) = \frac{\pi}{2} + 0 = \frac{\pi}{2} \]

Now, let's consider the behavior of $g(x)$ for $-1 < x < 1$. We can rewrite $g(x)$ as
\[ g(x) = \arcsin x + \arcsin \sqrt{1-x^2} = \arcsin x + \arcsin \sqrt{1-x^2} \]

Since $-1 < x < 1$, we know that $0 < \sqrt{1-x^2} < 1$. This means that the second term, $\arcsin \sqrt{1-x^2}$, will always be non-negative.

As $x$ increases from $-1$ to $1$, the first term, $\arcsin x$, increases from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. The second term, $\arcsin \sqrt{1-x^2}$, decreases from $\frac{\pi}{2}$ to $0$.

The sum of these two terms will result in a range that spans from $-\frac{\pi}{2} + \frac{\pi}{2} = 0$ to $\frac{\pi}{2} + 0 = \frac{\pi}{2}$.

However, we can simplify $g(x)$ further by using the identity $\arcsin \sqrt{1-x^2} = \frac{\pi}{2} - \arcsin x$. Then,
\[ g(x) = \arcsin x + \left( \frac{\pi}{2} - \arcsin x \right) = \frac{\pi}{2} \]

This simplification shows that $g(x)$ is actually a constant function, and its range is simply $\boxed{\frac{\pi}{2}}$.