`tan^(6)20^(@) - 33tan^(4) 20^(@) + 27 tan^(2) 20^(@) + 4=`

Answer

To solve the equation \( \tan^6(20^\circ) - 33\tan^4(20^\circ) + 27\tan^2(20^\circ) + 4 = 0 \), we can follow these steps:

### Step 1: Substitute \( x = \tan^2(20^\circ) \)
Let \( x = \tan^2(20^\circ) \). Then, we can rewrite the equation in terms of \( x \):
\[ x^3 - 33x^2 + 27x + 4 = 0 \]

### Step 2: Use the identity for \( \tan(3\theta) \)
We know that:
\[ \tan(3\theta) = \frac{3\tan(\theta) - \tan^3(\theta)}{1 - 3\tan^2(\theta)} \]
For \( \theta = 20^\circ \), we have \( 3\theta = 60^\circ \) and \( \tan(60^\circ) = \sqrt{3} \). Thus, we can set up the equation:
\[ \sqrt{3} = \frac{3\tan(20^\circ) - \tan^3(20^\circ)}{1 - 3\tan^2(20^\circ)} \]

### Step 3: Cross-multiply and simplify
Cross-multiplying gives:
\[ \sqrt{3}(1 - 3\tan^2(20^\circ)) = 3\tan(20^\circ) - \tan^3(20^\circ) \]
Substituting \( x = \tan^2(20^\circ) \) leads to:
\[ \sqrt{3}(1 - 3x) = 3\sqrt{x} - x^{3/2} \]

### Step 4: Square both sides
Squaring both sides to eliminate the square roots:
\[ 3(1 - 3x)^2 = (3\sqrt{x} - x^{3/2})^2 \]

### Step 5: Expand both sides
Expanding both sides gives:
\[ 3(1 - 6x + 9x^2) = 9x - 6x^2 + x^3 \]
This simplifies to:
\[ 3 - 18x + 27x^2 = 9x - 6x^2 + x^3 \]

### Step 6: Rearranging the equation
Rearranging all terms to one side results in:
\[ x^3 - 33x^2 + 27x + 4 = 0 \]

### Step 7: Solve the cubic equation
To find the roots of the cubic equation, we can use the Rational Root Theorem or synthetic division. Testing for rational roots, we find that \( x = 1 \) is a root.

### Step 8: Factor the cubic equation
Using synthetic division, we can factor the cubic equation:
\[ (x - 1)(x^2 - 32x - 4) = 0 \]

### Step 9: Solve the quadratic equation
Now, we can solve the quadratic equation \( x^2 - 32x - 4 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{32 \pm \sqrt{1024 + 16}}{2} = \frac{32 \pm \sqrt{1040}}{2} \]
This simplifies to:
\[ x = 16 \pm 2\sqrt{65} \]

### Step 10: Substitute back to find \( \tan^2(20^\circ) \)
Since \( x = \tan^2(20^\circ) \), we can now find the values of \( \tan^2(20^\circ) \) and, consequently, \( \tan(20^\circ) \).

### Final Result
The original equation evaluates to:
\[ \tan^6(20^\circ) - 33\tan^4(20^\circ) + 27\tan^2(20^\circ) + 4 = 7 \]