Bonjour,
1)
[tex]f(t)=0\\\\\dfrac{\sqrt{2}}{2}\times cos[\dfrac{1}{4}(t-\pi)] =0\\\\cos[\dfrac{1}{4}(t-\pi)] =0\\\\cos[\dfrac{1}{4}(t-\pi)] =cos(\dfrac{\pi}{2})\ \ \ car\ \ cos(\dfrac{\pi}{2})=0[/tex]
2) [tex]cos[\dfrac{1}{4}(t-\pi)] =cos(\dfrac{\pi}{2})\Longleftrightarrow \dfrac{1}{4}(t-\pi) =\dfrac{\pi}{2}+k\pi\ \ (k\in\matbb{Z})\\\\\Longleftrightarrow t-\pi =2\pi+4k\pi\ \ (k\in\matbb{Z})\\\\\Longleftrightarrow t =3\pi+4k\pi\ \ (k\in\matbb{Z})[/tex]
Si k = 0, alors [tex]t_1=3\pi\ secondes\approx 9,4\ secondes[/tex]
3) [tex]0\le t\le35\Longrightarrow 0\le 3\pi+4k\pi\le35\ \ (k\in\matbb{Z})\\\\\Longrightarrow -3\pi\le 4k\pi\le35-3\pi\ \ (k\in\matbb{Z})\\\\\Longrightarrow \dfrac{-3}{4}\le k\le\dfrac{35-3\pi}{4\pi}\approx2,03...\ \ (k\in\matbb{Z})\\\\\Longrightarrow -0,75\le k\le2,03...\ \ (k\in\matbb{Z})[/tex]
Puisque k est entier, nous aurons : k = 0 ; 1 et 2 ==> 3 fois.
4) graphique en pièce jointe.
