Rozwiązanie:
[tex]sin9x+sin5x=sin7x\\2sin\frac{9x+5x}{2} *cos\frac{9x-5x}{2} =sin7x\\2sin7x*cos2x=sin7x\\2sin7x*cos2x-sin7x=0\\sin7x(2cos2x-1)=0\\sin7x =0 \vee cos2x=\frac{1}{2} \\7x=k\pi \vee 2x=\frac{\pi }{3}+2k\pi \vee 2x=-\frac{\pi }{3}+2k\pi \\x=\frac{k\pi }{7} \vee x=\frac{\pi}{6} +k\pi \vee x=-\frac{\pi }{6}+k\pi \\k \in \mathbb{Z}[/tex]
