Question

Tg x - √3tg x +1= √3

Answer 1

tgx - √3•tgx + 1 = √3

tg•(1 - √3) + (1 - √3) = 0

(tgx + 1)(1 - √3) = 0

tgx + 1 = 0

tgx = - 1

x = (-π/4) + πn, n ∈ Z

ОТВЕТ: (-π/4) + πn, n ∈ Z

Answer 2

To solve the equation $\tan(x) - \sqrt{3}\tan(x) + 1 = \sqrt{3}$, you can follow these steps:

1. Combine like terms: $(1 - \sqrt{3})\tan(x) + 1 = \sqrt{3}$

2. Move the constant term (1) to the right side of the equation: $(1 - \sqrt{3})\tan(x) = \sqrt{3} - 1$

3. Divide both sides by $(1 - \sqrt{3})$ to isolate $\tan(x)$: $\tan(x) = \frac{\sqrt{3} - 1}{1 - \sqrt{3}}$

4. Simplify the right side of the equation: $\tan(x) = \frac{\sqrt{3} - 1}{1 - \sqrt{3}}$

5. Rationalize the denominator. To do this, multiply both the numerator and denominator by the conjugate of the denominator: $\tan(x) = \frac{(\sqrt{3} - 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})}$

6. Simplify further: $\tan(x) = \frac{\sqrt{3} - 1}{1 - 3}$

7. Continue to simplify: $\tan(x) = \frac{\sqrt{3} - 1}{-2}$

8. Finally, solve for $x$ by taking the arctan (inverse tangent) of both sides: $x = \arctan\left(\frac{\sqrt{3} - 1}{-2}\right)$

Now, you can use a calculator to find the value of $x$.