To solve this trigonometric equation, we can use the identity:
[ \text{ctg}x = \frac{1}{\text{tg}x} ]
So, the given equation can be rewritten as:
[ \text{tg}(x-15^\circ) \cdot \text{tg}(x+15^\circ) = 1/3 ]
We can also use the identity:
[ \text{tg}(A - B) = \frac{\text{tg}A - \text{tg}B}{1 + \text{tg}A \cdot \text{tg}B} ]
Applying this identity to the left side of the equation, we get:
[ \frac{\text{tg}x - \text{tg}15^\circ}{1 + \text{tg}x \cdot \text{tg}15^\circ} \cdot \frac{\text{tg}x + \text{tg}15^\circ}{1 - \text{tg}x \cdot \text{tg}15^\circ} = \frac{1}{3} ]
[ \frac{(\text{tg}x)^2 - (\text{tg}15^\circ)^2}{1 - (\text{tg}x)^2 \cdot (\text{tg}15^\circ)^2} = \frac{1}{3} ]
Since we know that $\text{tg}15^\circ = \sqrt{3} -2$, we can substitute this value into the equation. Then, we can solve for $\text{tg}x$ by solving this quadratic equation.
To solve this trigonometric equation, we can use the identity:
[ \text{ctg}x = \frac{1}{\text{tg}x} ]
So, the given equation can be rewritten as:
[ \text{tg}(x-15^\circ) \cdot \text{tg}(x+15^\circ) = 1/3 ]
We can also use the identity:
[ \text{tg}(A - B) = \frac{\text{tg}A - \text{tg}B}{1 + \text{tg}A \cdot \text{tg}B} ]
Applying this identity to the left side of the equation, we get:
[ \frac{\text{tg}x - \text{tg}15^\circ}{1 + \text{tg}x \cdot \text{tg}15^\circ} \cdot \frac{\text{tg}x + \text{tg}15^\circ}{1 - \text{tg}x \cdot \text{tg}15^\circ} = \frac{1}{3} ]
[ \frac{(\text{tg}x)^2 - (\text{tg}15^\circ)^2}{1 - (\text{tg}x)^2 \cdot (\text{tg}15^\circ)^2} = \frac{1}{3} ]
Since we know that $\text{tg}15^\circ = \sqrt{3} -2$, we can substitute this value into the equation. Then, we can solve for $\text{tg}x$ by solving this quadratic equation.