Odpowiedź:
[tex]$x \in \Big \{\frac{\pi}{4} ,\frac{5\pi}{4} ,\frac{4\pi}{3} \Big \}[/tex]
Szczegółowe wyjaśnienie:
Zauważmy, że:
[tex]$\sin\frac{x}{2} - \cos\frac{x}{2} =2 \sin\frac{x+x-\pi }{4} \cos\frac{x-x+\pi }{4} =2 \sin \Big(\frac{x}{2} -\frac{\pi}{4} \Big) \cos\frac{\pi}{4} =[/tex]
[tex]$=\sqrt{2} \sin \Big(\frac{x}{2} -\frac{\pi}{4} \Big)[/tex]
[tex]$\sin \frac{3}{2} x- \cos \frac{3}{2} x=2 \sin \frac{3x+3x-\pi }{4} \cos\frac{3x-3x+\pi }{4}=2\sin\Big(\frac{3}{2}x -\frac{\pi}{4} \Big) \cos\frac{\pi}{4} =[/tex]
[tex]$=\sqrt{2}\sin \Big(\frac{3}{2}x-\frac{\pi}{4} \Big)[/tex]
[tex]$\cos x-\sin x=-2 \sin\frac{2x+2x-\pi }{4} \cos \frac{2x-2x+\pi }{4} =-2\sin \Big(x -\frac{\pi}{4} \Big) \cos \frac{\pi }{4} =[/tex]
[tex]$=-\sqrt{2} \sin \Big(x-\frac{\pi}{4}\Big)[/tex]
Zatem równanie wygląda tak:
[tex]$\sqrt{2} \sin \Big(\frac{x}{2} -\frac{\pi}{4} \Big) + \sqrt{2}\sin \Big(\frac{3}{2}x-\frac{\pi}{4} \Big)=-\sqrt{2} \sin \Big(x-\frac{\pi}{4}\Big)[/tex]
[tex]$ \sin \Big(\frac{x}{2} -\frac{\pi}{4} \Big) +\sin \Big(\frac{3}{2}x-\frac{\pi}{4} \Big)+ \sin \Big(x-\frac{\pi}{4}\Big)=0[/tex]
Teraz mamy:
[tex]$ \sin \Big(\frac{x}{2} -\frac{\pi}{4} \Big) +\sin \Big(\frac{3}{2}x-\frac{\pi}{4} \Big)=2 \sin \frac{\frac{x}{2} -\frac{\pi}{4} +\frac{3}{2}x-\frac{\pi}{4} }{2} \cos \frac{\frac{x}{2} -\frac{\pi}{4} -\frac{3}{2}x+\frac{\pi}{4} }{2} =[/tex]
[tex]$=2 \sin \frac{2x-\frac{\pi}{2} }{2} \cos\frac{x}{2} =2 \sin\Big(x-\frac{\pi}{4} \Big) \cos\frac{x}{2}[/tex]
Zatem równanie wygląda tak:
[tex]$2 \sin\Big(x-\frac{\pi}{4} \Big) \cos\frac{x}{2}+ \sin \Big(x-\frac{\pi}{4}\Big)=0[/tex]
[tex]$ \sin \Big(x-\frac{\pi}{4}\Big) \Big(2 \cos \frac{x}{2} +1\Big)=0[/tex]
Stąd:
[tex]$\sin \Big(x-\frac{\pi}{4}\Big)=0 \vee 2 \cos\frac{x}{2} +1=0[/tex]
[tex]$ \sin \Big(x-\frac{\pi}{4}\Big)=0 \iff x-\frac{\pi}{4} =k\pi \iff x=\frac{\pi}{4}+k\pi[/tex]
[tex]$2\cos \frac{x}{2} +1=0 \iff \cos \frac{x}{2} =-\frac{1}{2} \iff \frac{x}{2} =\frac{2\pi}{3} +2k\pi \vee \frac{x}{2} =-\frac{2\pi}{3} +2k\pi[/tex]
[tex]$\iff x=\frac{4\pi}{3} +4k\pi \vee x=-\frac{4\pi}{3} +4k\pi[/tex]
Pozostało znaleźć rozwiązania w danym przedziale:
[tex]$x \in \Big \{\frac{\pi}{4} ,\frac{5\pi}{4} ,\frac{4\pi}{3} \Big \}[/tex]
