Question

Rezolvati ecuatia: 12sin(3x)-5cos(3x)=13

Answer 1

[tex]12\sin(3x)-5\cos(3x)=13\\ 12\sin(3x)=13+5\cos(3x)|()^2\\ 144\sin^2(3x)=169+130\cos(3x)+25\cos^2(3x)\\ 144(1-\cos^2(3x))=169+130\cos(3x)+25\cos^2(3x)\\ 144-144\cos^2(3x)=169+130\cos(3x)+25\cos^2 (3x)\\ 169\cos^2(3x)+130\cos(3x)+25=0\\ (13\cos(3x)+5)^2=0\\ 13\cos(3x)+5=0\\ \cos(3x)=-\dfrac{5}{13}\\ 3x=\pm \arccos(-\dfrac{5}{13})+2k\pi,k\in \mathbb{Z}\\ x=\dfrac{1}{3}\left(\pm \arccos(-\dfrac{5}{13})+2k\pi\right),k\in \mathbb{Z}[/tex]

Answer 2

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