Question

(sinx+cosx)/(sinx-cosx)>√3​

Answer 1

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

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

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

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

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

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

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

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

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

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

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