Find the number of solutions for `sinxtan4x=cosx`, when `x in (0,pi)`

Answer

To solve the equation \( \sin x \tan 4x = \cos x \) for \( x \) in the interval \( (0, \pi) \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the original equation:
\[ \sin x \tan 4x = \cos x \]
We know that \( \tan 4x = \frac{\sin 4x}{\cos 4x} \). Thus, we can rewrite the equation as:
\[ \sin x \cdot \frac{\sin 4x}{\cos 4x} = \cos x \]
Multiplying both sides by \( \cos 4x \) (assuming \( \cos 4x \neq 0 \)):
\[ \sin x \sin 4x = \cos x \cos 4x \]

### Step 2: Use the product-to-sum identities
Using the product-to-sum identities, we can rewrite both sides:
\[ \sin x \sin 4x = \frac{1}{2} [\cos (x - 4x) - \cos (x + 4x)] = \frac{1}{2} [\cos (-3x) - \cos (5x)] \]
\[ \cos x \cos 4x = \frac{1}{2} [\cos (x + 4x) + \cos (x - 4x)] = \frac{1}{2} [\cos (5x) + \cos (-3x)] \]
Setting these equal gives:
\[ \frac{1}{2} [\cos (-3x) - \cos (5x)] = \frac{1}{2} [\cos (5x) + \cos (-3x)] \]

### Step 3: Simplify the equation
Multiplying through by 2 to eliminate the fraction:
\[ \cos (-3x) - \cos (5x) = \cos (5x) + \cos (-3x) \]
Rearranging terms:
\[ \cos (-3x) - \cos (-3x) = 2\cos (5x) \]
This simplifies to:
\[ 0 = 2\cos (5x) \]
Thus:
\[ \cos (5x) = 0 \]

### Step 4: Solve for \( x \)
The general solution for \( \cos (5x) = 0 \) is:
\[ 5x = \frac{\pi}{2} + k\pi \quad \text{for } k \in \mathbb{Z} \]
This simplifies to:
\[ x = \frac{\pi}{10} + \frac{k\pi}{5} \]

### Step 5: Find the values of \( k \)
Now we need to find the values of \( k \) such that \( x \) is in the interval \( (0, \pi) \):
1. For \( k = 0 \):
\[ x = \frac{\pi}{10} \approx 0.314 \quad (\text{valid}) \]
2. For \( k = 1 \):
\[ x = \frac{\pi}{10} + \frac{\pi}{5} = \frac{3\pi}{10} \approx 0.942 \quad (\text{valid}) \]
3. For \( k = 2 \):
\[ x = \frac{\pi}{10} + \frac{2\pi}{5} = \frac{5\pi}{10} = \frac{\pi}{2} \quad (\text{valid}) \]
4. For \( k = 3 \):
\[ x = \frac{\pi}{10} + \frac{3\pi}{5} = \frac{7\pi}{10} \approx 2.199 \quad (\text{valid}) \]
5. For \( k = 4 \):
\[ x = \frac{\pi}{10} + \frac{4\pi}{5} = \frac{9\pi}{10} \approx 2.827 \quad (\text{valid}) \]
6. For \( k = 5 \):
\[ x = \frac{\pi}{10} + \frac{5\pi}{5} = \frac{11\pi}{10} \quad (\text{not valid, as it exceeds } \pi) \]

### Conclusion
Thus, the valid values of \( k \) are \( 0, 1, 2, 3, 4 \), giving us a total of **5 solutions** in the interval \( (0, \pi) \).