The equation `2sin(x/2).cos^2x+sin^2x=2sin(x/2)sin^2x+cos^2x` has a root for which

Answer

To solve the equation \( 2\sin\left(\frac{x}{2}\right)\cos^2 x + \sin^2 x = 2\sin\left(\frac{x}{2}\right)\sin^2 x + \cos^2 x \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ 2\sin\left(\frac{x}{2}\right)\cos^2 x + \sin^2 x = 2\sin\left(\frac{x}{2}\right)\sin^2 x + \cos^2 x \]
We can rearrange this to group similar terms:
\[ 2\sin\left(\frac{x}{2}\right)\cos^2 x - 2\sin\left(\frac{x}{2}\right)\sin^2 x = \cos^2 x - \sin^2 x \]

### Step 2: Factoring Out Common Terms
Notice that we can factor out \( 2\sin\left(\frac{x}{2}\right) \) from the left-hand side:
\[ 2\sin\left(\frac{x}{2}\right)(\cos^2 x - \sin^2 x) = \cos^2 x - \sin^2 x \]

### Step 3: Setting Up the Equation
Now, we can set up the equation:
\[ 2\sin\left(\frac{x}{2}\right)(\cos^2 x - \sin^2 x) - (\cos^2 x - \sin^2 x) = 0 \]
Factoring gives us:
\[ (\cos^2 x - \sin^2 x)(2\sin\left(\frac{x}{2}\right) - 1) = 0 \]

### Step 4: Solving Each Factor
Now we have two factors to solve:
1. \( \cos^2 x - \sin^2 x = 0 \)
2. \( 2\sin\left(\frac{x}{2}\right) - 1 = 0 \)

**For the first factor:**
\[ \cos^2 x = \sin^2 x \implies \tan^2 x = 1 \implies \tan x = \pm 1 \]
Thus, \( x = n\pi + \frac{\pi}{4} \) or \( x = n\pi + \frac{3\pi}{4} \), where \( n \) is an integer.

**For the second factor:**
\[ 2\sin\left(\frac{x}{2}\right) = 1 \implies \sin\left(\frac{x}{2}\right) = \frac{1}{2} \]
This gives:
\[ \frac{x}{2} = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad \frac{x}{2} = \frac{5\pi}{6} + 2k\pi \]
Thus, \( x = \frac{\pi}{3} + 4k\pi \) or \( x = \frac{5\pi}{3} + 4k\pi \).

### Step 5: Conclusion
The roots of the original equation are given by the solutions from both factors:
- From \( \tan x = \pm 1 \): \( x = n\pi + \frac{\pi}{4} \) or \( x = n\pi + \frac{3\pi}{4} \)
- From \( \sin\left(\frac{x}{2}\right) = \frac{1}{2} \): \( x = \frac{\pi}{3} + 4k\pi \) or \( x = \frac{5\pi}{3} + 4k\pi \)