Question

aratati ca 2√2 tgx +1=0,stiind ca sin x=1/3 si x apartine (pi/2,pi)

Answer 1

$\it x\in \left(\dfrac{\pi}{2}, \pi \right) \Rightarrow cosx \ \textless \ 0\\ \\ \\ cosx= -\sqrt{1- sin^2x} =-\sqrt{1-\dfrac{1}{9}} =-\sqrt{\dfrac{8}{9}} =-\dfrac{2\sqrt2}{3}$

[tex]\it 2\sqrt2 tgx +1= 2\sqrt2\dfrac{sinx}{cosx} +1 = 2\sqrt2\cdot\dfrac{\dfrac{1}{3}}{-\dfrac{2\sqrt2}{3}} +1 = \\ \\ \\ = -2\sqrt2\cdot\dfrac{1}{3}\cdot\dfrac{3}{2\sqrt2} +1 = -1+1=0[/tex]