https://file.aiquickdraw.com/mathbot/file/1726694547414hro4vayh.jpg

The equation in the image is:

\[
\arcsin(2x) = \arccos(x)
\]

To solve this equation, we can use the relationship between the sine and cosine functions:

\[
\arcsin(y) + \arccos(y) = \frac{\pi}{2}
\]

So, for the given equation, we can rewrite it as:

\[
\arcsin(2x) + \arccos(x) = \frac{\pi}{2}
\]

Substituting \(\arccos(x)\) from the original equation:

\[
\arcsin(2x) + \arcsin(2x) = \frac{\pi}{2}
\]

This simplifies to:

\[
2\arcsin(2x) = \frac{\pi}{2}
\]

Now, divide both sides by 2:

\[
\arcsin(2x) = \frac{\pi}{4}
\]

Next, we apply the sine function to both sides:

\[
2x = \sin\left(\frac{\pi}{4}\right)
\]

Since \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), we get:

\[
2x = \frac{\sqrt{2}}{2}
\]

Finally, divide by 2 to solve for \(x\):

\[
x = \frac{\sqrt{2}}{4}
\]

So, the solution is:

\[
x = \frac{\sqrt{2}}{4}
\]

Would you like further details or clarifications on this solution?

### 5 Related Questions:
1. What are the general properties of inverse trigonometric functions?
2. How do we derive the identity \(\arcsin(y) + \arccos(y) = \frac{\pi}{2}\)?
3. Can you apply the same method to solve \(\arcsin(x) = \arccos(2x)\)?
4. What happens if the value of \(x\) is outside the range of the inverse sine and cosine functions?
5. How do you convert between inverse sine and cosine expressions?

### Tip:
When solving inverse trigonometric equations, remember to always check if the solution lies within the defined range of the inverse function, such as \(-1 \leq x \leq 1\) for arcsine and arccosine functions.

Learn how to solve the trigonometric equation arcsin(2x) = arccos(x) using inverse trigonometric identities. This problem explores key concepts such as inverse sine and cosine functions, with a detailed step-by-step breakdown. Suitable for high school students in Grades 10-12.

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