Question

Exercițiul 9 va rog mult. Răspunsul ar trebui sa dea rad7-2

Answer 1

$\sin x = \dfrac{2\cdot \text{tg}\dfrac{x}{2}}{1+\text{tg}^2\dfrac{x}{2}};\quad \cos x = \dfrac{1-\text{tg}^2\dfrac{x}{2}}{1+\text{tg}^2\dfrac{x}{2}} \\ \\ Notam \text{tg}\dfrac{x}{2} = t;\\ \\ x\in \Big(0,\dfrac{\pi}{2}\Big)\Big|:2 \Rightarrow \dfrac{x}{2}\in \Big(0,\dfrac{\pi}{4}\Big) \Rightarrow \text{tg}\dfrac{x}{2} \in \Big(\text{tg} 0,\text{tg}\dfrac{\pi}{4}\Big) \Rightarrow \text{tg}\dfrac{x}{2} \in (0,1)\Rightarrow \\ \\ \Rightarrow t\in (0,1)\\ \\ Trecem la expresie:$
$\sin x-\cos x = \dfrac{1}{2} \Rightarrow \dfrac{2t}{1+t^2}-\dfrac{1-t^2}{1+t^2}=\dfrac{1}{2} \Rightarrow \dfrac{2t-(1-t^2)}{1+t^2} = \dfrac{1}{2} \Rightarrow \\ \\ \Rightarrow \dfrac{t^2+2t-1}{1+t^2} = \dfrac{1}{2} \Rightarrow 2\cdot(t^2+2t-1) = 1+t^2 \Rightarrow \\ \\ \Rightarrow 2t^2+4t-2=1+t^2 \Rightarrow t^2+4t-3 = 0\\ \\ \Delta = 16+12 = 28 \Rightarrow t_{1,2} = \dfrac{-4\pm \sqrt{28}}{2} \Rightarrow t_{1,2} = \dfrac{-4\pm 2\sqrt7}{2} \Rightarrow \\ \\ \Rightarrow t_{1,2} = -2\pm\sqrt7$

$\bullet t_1 = -2-\sqrt7 \notin (0,1) \\ \bullet t_2 = -2+\sqrt7 \in (0,1) \\ \\ \Rightarrow t = \sqrt7-2 \Rightarrow \text{tg}\dfrac{x}{2} = \sqrt7-2 \Rightarrow \boxed{S = \Big\{\sqrt7-2\Big\}}$