The equation you've provided is:

\[
2y + \cos \left( \frac{1}{y} \right) - 2y = 2x
\]

Simplifying the equation:

\[
\cancel{2y} + \cos \left( \frac{1}{y} \right) - \cancel{2y} = 2x
\]

This simplifies to:

\[
\cos \left( \frac{1}{y} \right) = 2x
\]

Now, to solve for \( y \), we take the inverse cosine (arccos) of both sides:

\[
\frac{1}{y} = \arccos(2x)
\]

Finally, solving for \( y \):

\[
y = \frac{1}{\arccos(2x)}
\]

So, the solution is:

\[
y = \frac{1}{\arccos(2x)}
\]

Do you want further details or have any questions?

Here are 5 related questions:
1. What happens if \( 2x \) is outside the domain of the cosine function?
2. How do we find the domain of the function \( y = \frac{1}{\arccos(2x)} \)?
3. Can this equation have complex solutions?
4. How would you graph the solution for various values of \( x \)?
5. What are the limits of \( y \) as \( x \) approaches certain values?

**Tip:** Remember that the arccos function is defined only for values between -1 and 1. Always check the domain when working with trigonometric functions.

Discover how to simplify and solve the equation 2y + cos(1/y) - 2y = 2x using trigonometric functions. This problem focuses on inverse trigonometric concepts and is ideal for high school students tackling advanced algebra and trigonometry. Learn how to manipulate the cosine and arccosine functions to find y.

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