Odpowiedź :
Odpowiedź:
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Proszę bardzo! :)
[tex]cos(\frac{\pi }{4}-2x)=cos(x+\frac{\pi }{3})\\\\cos(\frac{\pi }{4}-2x)-cos(x+\frac{\pi }{3})=0[/tex]
Skorzystajmy z różnicy funkcji trygonometrycznych:
[tex]cos\alpha -cos\beta =-2sin\frac{\alpha +\beta }{2}*sin\frac{\alpha -\beta }{2}[/tex]
Nasze α to:
[tex]\frac{\pi }{4}-2x[/tex]
β to:
[tex]x+\frac{\pi }{3}[/tex]
Podstawiamy!
[tex]cos(\frac{\pi }{4}-2x)-cos(x+\frac{\pi }{3})=-2sin\frac{\frac{\pi }{4}-2x+x+\frac{\pi }{3} }{2} *sin\frac{\frac{\pi }{4}-2x-x-\frac{\pi }{3} }{2}[/tex]
[tex]-2sin\frac{-x+\frac{3\pi }{12}+\frac{4\pi }{12} }{2}*sin\frac{-3x+\frac{3\pi }{12}-\frac{4\pi }{12} }{2}[/tex]
[tex]-2sin(\frac{\frac{7\pi }{12}-x }{2})sin(\frac{\frac{-\pi }{12}-3x }{2})=0 \\\\-2sin(\frac{\frac{7\pi }{12}-\frac{12x}{12} }{2})sin(\frac{\frac{-\pi }{12}-\frac{36x }{12} }{2})=0 \\\\-2sin(\frac{\frac{7\pi-12x }{12} }{2})sin(\frac{\frac{-\pi -36x }{12} }{2})=0\\\\ -2sin(\frac{7\pi -12x}{12}*\frac{1}{2})sin(\frac{-\pi -36x}{12}*\frac{1}{2})=0\\\\-2sin(\frac{7\pi -12x}{24})sin(\frac{-\pi -36x}{24})=0\ \ \ /:(-2)\\\\ sin(\frac{7\pi -12x}{24})sin(\frac{-\pi-36x}{24})=0\\\\[/tex]
1)
[tex]sin(\frac{7\pi -12x}{24})=0\\\\\frac{7\pi -12x}{24}=0\ \ \ /*24\\\\7\pi -12x=0\\\\-12x=-7\pi\ \ /:(-12)\\\\x=\frac{7\pi }{12}+2k\pi[/tex], k∈C
2)
[tex]sin(\frac{-\pi -36x}{24})=0\\\\\frac{-\pi -36x}{24}=0\ \ \ /*(24)\\\\-\pi -36x=0\\\\-\pi -36x=0\\\\-36x =\pi \ \ \ /:(-36)\\\\x=-\frac{\pi }{36}+2k\pi[/tex], k∈C
Szczegółowe wyjaśnienie:
