The real roots of the equation `cos^7x+sin^4x=1` in the interval `(-pi,pi)` are __________, ________, and ________

Explanation

To solve the equation \( \cos^7 x + \sin^4 x = 1 \) for real roots in the interval \( (-\pi, \pi) \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos^7 x + \sin^4 x = 1 \]
We can rearrange this to isolate \( \cos^7 x \):
\[ \cos^7 x = 1 - \sin^4 x \]

### Step 2: Express \( \sin^4 x \) in terms of \( \cos^2 x \)
Recall that \( \sin^2 x = 1 - \cos^2 x \). Thus, we can express \( \sin^4 x \) as:
\[ \sin^4 x = (1 - \cos^2 x)^2 = 1 - 2\cos^2 x + \cos^4 x \]
Substituting this back into the equation gives:
\[ \cos^7 x = 1 - (1 - 2\cos^2 x + \cos^4 x) \]
Simplifying this, we have:
\[ \cos^7 x = 2\cos^2 x - \cos^4 x \]

### Step 3: Rearranging the equation
Rearranging the equation yields:
\[ \cos^7 x + \cos^4 x - 2\cos^2 x = 0 \]
Let \( y = \cos^2 x \). Then, we can rewrite the equation as:
\[ y^3 + y^2 - 2y = 0 \]

### Step 4: Factor the polynomial
Factoring out \( y \) gives:
\[ y(y^2 + y - 2) = 0 \]
Now, we can factor \( y^2 + y - 2 \):
\[ y^2 + y - 2 = (y - 1)(y + 2) \]
Thus, the complete factorization is:
\[ y(y - 1)(y + 2) = 0 \]

### Step 5: Solve for \( y \)
Setting each factor to zero gives:
1. \( y = 0 \) → \( \cos^2 x = 0 \) → \( \cos x = 0 \)
2. \( y - 1 = 0 \) → \( \cos^2 x = 1 \) → \( \cos x = \pm 1 \)
3. \( y + 2 = 0 \) → Not valid since \( y = \cos^2 x \) cannot be negative.

### Step 6: Find the values of \( x \)
1. From \( \cos x = 0 \):
- \( x = \frac{\pi}{2}, -\frac{\pi}{2} \)

2. From \( \cos x = 1 \):
- \( x = 0 \)

3. From \( \cos x = -1 \):
- \( x = \pi, -\pi \) (but \(-\pi\) is equivalent to \(\pi\) in the interval \((- \pi, \pi)\))

### Step 7: List the roots
The real roots of the equation \( \cos^7 x + \sin^4 x = 1 \) in the interval \( (-\pi, \pi) \) are:
\[ -\frac{\pi}{2}, 0, \frac{\pi}{2} \]

### Final Answer
The real roots of the equation in the interval \( (-\pi, \pi) \) are:
\[ -\frac{\pi}{2}, 0, \frac{\pi}{2} \]