Solve `(sin 10^(@))^(tan x+tan 3x)=tan 15^(@)+tan 30^(@)+ tan 15^(@). Tan 30^(@), x in (0, pi]`.

Explanation

To solve the equation \((\sin 10^\circ)^{(\tan x + \tan 3x)} = \tan 15^\circ + \tan 30^\circ + \tan 15^\circ \cdot \tan 30^\circ\) for \(x\) in the interval \((0, \pi]\), we can follow these steps:

### Step 1: Simplify the Right-Hand Side (RHS)
First, we need to calculate the values of \(\tan 15^\circ\) and \(\tan 30^\circ\).

Using the tangent addition formula:
\[ \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \]
Where \(\tan 45^\circ = 1\) and \(\tan 30^\circ = \frac{1}{\sqrt{3}}\):
\[ \tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

Now, calculate \(\tan 30^\circ\):
\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

Now substituting these values into the RHS:
\[ \tan 15^\circ + \tan 30^\circ + \tan 15^\circ \cdot \tan 30^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} + \frac{1}{\sqrt{3}} + \left(\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cdot \frac{1}{\sqrt{3}}\right) \]

### Step 2: Calculate the Product
Calculating the product:
\[ \tan 15^\circ \cdot \tan 30^\circ = \frac{(\sqrt{3} - 1)}{(\sqrt{3} + 1) \cdot \sqrt{3}} = \frac{\sqrt{3} - 1}{3 + \sqrt{3}} \]

Now, combine all terms:
\[ \text{RHS} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} + \frac{1}{\sqrt{3}} + \frac{\sqrt{3} - 1}{3 + \sqrt{3}} \]

### Step 3: Solve for the Left-Hand Side (LHS)
The LHS is \((\sin 10^\circ)^{(\tan x + \tan 3x)}\). For this to equal the RHS, we need to find when \(\tan x + \tan 3x = 0\).

Using the identity:
\[ \tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} \]
Setting \(\tan x + \tan 3x = 0\):
\[ \tan x + \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} = 0 \]

### Step 4: Find Values of \(x\)
This leads to:
\[ \tan x (1 - 3\tan^2 x) + (3\tan x - \tan^3 x) = 0 \]
This simplifies to:
\[ \tan x (4 - 4\tan^2 x) = 0 \]
Thus, \(\tan x = 0\) or \(4 - 4\tan^2 x = 0\).

From \(\tan x = 0\), we have:
\[ x = n\pi \quad (n \in \mathbb{Z}) \]
In the interval \((0, \pi]\), we have \(x = \pi\).

From \(4 - 4\tan^2 x = 0\):
\[ \tan^2 x = 1 \Rightarrow \tan x = \pm 1 \]
This gives:
\[ x = \frac{\pi}{4}, \frac{3\pi}{4} \]

### Step 5: Final Values
Thus the solutions for \(x\) in the interval \((0, \pi]\) are:
\[ x = \frac{\pi}{4}, \frac{3\pi}{4}, \pi \]