Question

Ceva nu îmi dă bine…..ex 8

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\bf x\in\Big(\pi,\ \dfrac{3\pi}{2}\Big) \Rightarrow sinx < 0,\ \ cosx < 0\\ \\ \\ sinx=-\sqrt{1-cos^2x}=-\sqrt{1-\dfrac{8}{9}}=-\sqrt{\dfrac{1}{9}}=-\dfrac{1}{3}\\ \\ \\ tgx=\dfrac{sinx}{cosx}=\dfrac{-\dfrac{1}{3}}{-\dfrac{2\sqrt2}{3}}=\dfrac{^{\sqrt2)}1}{\ \ 2\sqrt2}=\dfrac{\sqrt2}{4}\\ \\ \\ R\breve aspunsul\ corect\ este\ B.\ \dfrac{\sqrt2}{4}$