Determine all solutions to the equation radical2 cos2x=sin^2x+cos^2x on the interval (0, 2pi)

The given equation is:

\[
\sqrt{2} \cos(2x) = \sin^2(x) + \cos^2(x)
\]

We know that:

\[
\sin^2(x) + \cos^2(x) = 1
\]

So the equation simplifies to:

\[
\sqrt{2} \cos(2x) = 1
\]

Next, solve for \( \cos(2x) \):

\[
\cos(2x) = \frac{1}{\sqrt{2}}
\]

The cosine of \( 2x \) equals \( \frac{1}{\sqrt{2}} \) at angles where \( 2x = \frac{\pi}{4} + 2k\pi \) or \( 2x = \frac{7\pi}{4} + 2k\pi \), where \( k \) is an integer.

Thus:

\[
2x = \frac{\pi}{4} + 2k\pi \quad \text{or} \quad 2x = \frac{7\pi}{4} + 2k\pi
\]

Now, solve for \( x \) by dividing by 2:

\[
x = \frac{\pi}{8} + k\pi \quad \text{or} \quad x = \frac{7\pi}{8} + k\pi
\]

Since we need to find the solutions in the interval \( (0, 2\pi) \), we consider \( k = 0 \) and \( k = 1 \):

1. For \( k = 0 \):
   - \( x = \frac{\pi}{8} \)
   - \( x = \frac{7\pi}{8} \)

2. For \( k = 1 \):
   - \( x = \frac{\pi}{8} + \pi = \frac{9\pi}{8} \)
   - \( x = \frac{7\pi}{8} + \pi = \frac{15\pi}{8} \)

Therefore, the solutions on the interval \( (0, 2\pi) \) are:

\[
x = \frac{\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{15\pi}{8}
\]

Would you like any details or have any further questions?

Here are some related questions you might find interesting:
1. How would the solutions change if the interval was \( [0, 4\pi] \)?
2. What is the general solution to the equation \( \cos(2x) = \frac{1}{\sqrt{2}} \)?
3. How would the problem change if \( \sin^2(x) + \cos^2(x) \) was a different constant?
4. What would the solutions be if the equation was \( \sqrt{2} \sin(2x) = 1 \) instead?
5. How would the equation change if we added a phase shift to \( \cos(2x) \)?

**Tip:** Remember that \( \cos(2x) \) and \( \sin(2x) \) represent periodic functions, so be sure to consider all relevant solutions within the given interval.

Explore how to solve the equation √2 cos(2x) = sin^2(x) + cos^2(x) on the interval (0, 2π). This problem involves trigonometric identities and the double-angle formula for cosine. Suitable for advanced high school students and those preparing for competitive exams.

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