Solve `cos 2x=|sin x|, x in (-pi/2, pi)`.

Explanation

To solve the equation \( \cos 2x = |\sin x| \) for \( x \) in the interval \( \left(-\frac{\pi}{2}, \pi\right) \), we can follow these steps:

### Step 1: Rewrite the equation
We know that \( \cos 2x \) can be expressed in terms of sine:
\[ \cos 2x = 2\cos^2 x - 1 = 2\sin^2 x - 1 \]
Thus, we can rewrite the equation as:
\[ 2\cos^2 x - 1 = |\sin x| \]

### Step 2: Consider the cases for \( |\sin x| \)
Since \( |\sin x| \) can be either \( \sin x \) or \( -\sin x \) depending on the value of \( x \), we will consider two cases:

**Case 1:** \( \sin x \geq 0 \) (which occurs in the intervals \( [0, \frac{\pi}{2}] \))
\[ \cos 2x = \sin x \]

**Case 2:** \( \sin x < 0 \) (which occurs in the intervals \( \left(-\frac{\pi}{2}, 0\right) \))
\[ \cos 2x = -\sin x \]

### Step 3: Solve Case 1
For \( \cos 2x = \sin x \):
\[ 2\cos^2 x - 1 = \sin x \]
Using \( \sin x = \sqrt{1 - \cos^2 x} \), we can substitute \( \sin x \) into the equation:
\[ 2\cos^2 x - 1 = \sqrt{1 - \cos^2 x} \]
Let \( \cos x = y \):
\[ 2y^2 - 1 = \sqrt{1 - y^2} \]
Squaring both sides:
\[ (2y^2 - 1)^2 = 1 - y^2 \]
Expanding and simplifying:
\[ 4y^4 - 4y^2 + 1 = 1 - y^2 \]
\[ 4y^4 - 3y^2 = 0 \]
Factoring out \( y^2 \):
\[ y^2(4y^2 - 3) = 0 \]
This gives us:
\[ y^2 = 0 \quad \text{or} \quad 4y^2 - 3 = 0 \]
From \( 4y^2 - 3 = 0 \):
\[ y^2 = \frac{3}{4} \quad \Rightarrow \quad y = \pm\frac{\sqrt{3}}{2} \]
Thus, \( \cos x = 0 \) or \( \cos x = \frac{\sqrt{3}}{2} \).

### Step 4: Find \( x \) values for Case 1
1. \( \cos x = 0 \) gives \( x = \frac{\pi}{2} \) (not in the interval).
2. \( \cos x = \frac{\sqrt{3}}{2} \) gives \( x = \frac{\pi}{6} \).

### Step 5: Solve Case 2
For \( \cos 2x = -\sin x \):
\[ 2\cos^2 x - 1 = -\sin x \]
This leads to:
\[ 2\cos^2 x - 1 = -\sqrt{1 - \cos^2 x} \]
Let \( \cos x = y \):
\[ 2y^2 - 1 = -\sqrt{1 - y^2} \]
Squaring both sides:
\[ (2y^2 - 1)^2 = 1 - y^2 \]
Following the same steps as before, we get:
\[ 4y^4 - 4y^2 + 1 = 1 - y^2 \]
\[ 4y^4 - 3y^2 = 0 \]
This gives us:
\[ y^2 = 0 \quad \text{or} \quad 4y^2 - 3 = 0 \]
From \( y^2 = 0 \), \( \cos x = 0 \) gives \( x = \frac{\pi}{2} \) (not in the interval).

From \( 4y^2 - 3 = 0 \):
\[ y^2 = \frac{3}{4} \quad \Rightarrow \quad y = -\frac{\sqrt{3}}{2} \]
Thus, \( \cos x = -\frac{\sqrt{3}}{2} \) gives \( x = \frac{5\pi}{6} \).

### Final Solutions
The solutions in the interval \( \left(-\frac{\pi}{2}, \pi\right) \) are:
\[ x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6} \]