Réponse :
1) On a
[tex]2cos(x+\frac{\pi }{3})=[/tex] [tex]2(cos(x)cos(\pi /3)-sin(x)sin(\frac{\pi }{3} ))[/tex](
= [tex]2(\frac{1}{2}cosx-\frac{\sqrt{3} }{2} sinx)[/tex]
=[tex]cosx-\sqrt{3}sinx[/tex]
=f(x)
2)
a)
[tex]cos(3x)-\sqrt{3}sin(3x)=\sqrt{3}[/tex]
⇔[tex]f(3x)=\sqrt{3}[/tex]
⇔[tex]2(cos(3x+\frac{\pi }{3} ))=\sqrt{3}[/tex]
⇔[tex]cos(3x+\frac{\pi }{3})=\frac{\sqrt{3} }{2}[/tex]
⇔[tex]3x+\frac{\pi }{3} = \frac{\pi }{6}[2\pi][/tex] ou [tex]3x+\frac{\pi }{3}= -\frac{\pi }{6}[2\pi ][/tex]
⇔
[tex]x=\frac{-\pi }{18}[2\pi ][/tex] ou [tex]x=-\frac{\pi }{6}[2\pi ][/tex]
b)
[tex]f(x)\geq -\sqrt{2}[/tex]
⇔[tex]2(cos(x+\frac{\pi }{3})\geq -\sqrt{2}[/tex]
⇔[tex]cos(x+\frac{\pi }{3})\geq -\frac{\sqrt{2} }{2}[/tex] car 2>0
⇔ [tex]0\leq x+\frac{\pi }{3} \leq \frac{3\pi }{4}[/tex] et [tex]\frac{5\pi }{4}\leq x+\frac{\pi }{3} \leq 2\pi[/tex]
⇔[tex]-\frac{\pi }{3}\leq x\leq \frac{5\pi }{12}[/tex] et [tex]\frac{11\pi }{12}\leq x \leq \frac{5\pi }{3}[/tex]
