Question

Exercitiul 9 iar raspunsul e d) 2 dar nu stiu cum

Answer 1

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Answer 2

$\it arcsin(cosx) = \dfrac{x}{2}\ \ \ \ \ (1) \\ \\ \\ (1) \Rightarrow cosx\in[-1,\ 1] \Rightarrow x\in[-\pi,\ \pi] \Rightarrow \dfrac{x}{2}\in \Big[-\dfrac{\pi}{2},\ \dfrac{\pi}{2}\Big]\ \ \ \ \ (2)\\ \\ \\ (1) \Rightarrow sin[arcsin(cosx)]=sin\dfrac{x}{2} \Rightarrow cosx=sin\dfrac{x}{2} \Rightarrow 1-2sin^2\dfrac{x}{2}=sin\dfrac{x}{2}$

$\it Not\breve{a}m\ \ \ sin\dfrac{x}{2} =t,\ iar\ ecua\c{\it t}ia\ devine:\\ \\ 1-2t^2=t \Rightarrow2t^2+t-1=0 \Rightarrow \begin{cases}\it t=-1\\ \\ sau\\ \\ \it t=\dfrac{1}{2} \end{cases}$

$\it t=-1 \Rightarrow sin\dfrac{x}{2} =-1 \stackrel{(2)}{\Longrightarrow}\ \dfrac{x}{2}=-\dfrac{\pi}{2} \Rightarrow x=-\pi\\ \\ \\ t=\dfrac{1}{2} \Rightarrow sin\dfrac{x}{2} =\dfrac{1}{2} \stackrel{(2)}{\Longrightarrow}\ \dfrac{x}{2}=\dfrac{\pi}{6} \Rightarrow x=\dfrac{\pi}{3}$

Mulțimea soluțiilor ecuației date este :

$\it S=\Big\{-\pi,\ \ \dfrac{\pi}{3}\Big\} \Rightarrow\ \ card(S) =2$