Question

(sinx+cosx)/(sinx-cosx)>√3​

Answer 1

$\dfrac{\sin x + \cos x}{\sin x - \cos x} > \sqrt{3}$

$\dfrac{\cos x \left(\dfrac{\sin x}{\cos x} + 1 \right)}{\cos x \left(\dfrac{\sin x}{\cos x} - 1 \right)} > \sqrt{3}$

$\dfrac{\text{tg} \, x + 1}{\text{tg} \, x - 1} > \sqrt{3}$

$-\dfrac{\text{tg} \, x + 1}{1 -\text{tg} \, x} > \sqrt{3}$

$-\dfrac{\text{tg} \, x + \text{tg} \, \dfrac{\pi}{4} }{1 -\text{tg}\, \dfrac{\pi}{4} \text{tg} \, x} > \sqrt{3}$

$-\text{tg} \left(x + \dfrac{\pi}{4} \right) > \sqrt{3}$

$\text{tg} \left(x + \dfrac{\pi}{4} \right) < -\sqrt{3}$

$-\dfrac{\pi}{2} + \pi n < x + \dfrac{\pi}{4} < \text{arctg}(-\sqrt{3}) + \pi n, \ n \in Z$

$-\dfrac{\pi}{2} + \pi n < x + \dfrac{\pi}{4} < -\dfrac{\pi}{3} + \pi n, \ n \in Z \ \ \ \ \ \ \ \bigg| - \dfrac{\pi}{4}$

$-\dfrac{3\pi}{4} + \pi n < x < -\dfrac{7\pi}{12} + \pi n, \ n \in Z$

Ответ: $-\dfrac{3\pi}{4} + \pi n < x < -\dfrac{7\pi}{12} + \pi n, \ n \in Z$