Rozwiązanie:
[tex]cosx+sin3x=0[/tex]
Zauważmy, że:
[tex]cosx=sin(\frac{ \pi }{2}-x)[/tex]
Zatem mamy:
[tex]sin(\frac{\pi }{2}-x)+sin3x=0[/tex]
Teraz korzystamy ze wzoru:
[tex]sin\alpha +sin\beta =2sin\frac{\alpha +\beta }{2}cos\frac{\alpha -\beta }{2}[/tex]
Dostaniemy:
[tex]2sin\frac{\frac{\pi }{2}-x+3x}{2}cos\frac{\frac{\pi }{2}-x-3x}{2}=0\\sin(x+\frac{\pi }{4} ) cos(\frac{\pi }{2}-2x)=0\\ sin(x+\frac{\pi }{4} )=0 \vee cos(\frac{\pi }{4}-2x)=0\\x+\frac{\pi }{4} =k\pi \vee \frac{\pi }{4}-2x=\frac{\pi }{2}+k\pi \\x=-\frac{\pi }{4}+k\pi \vee x=-\frac{\pi }{8} -\frac{k\pi }{2} \\k \in \mathbb{Z}[/tex]
