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Муодилаи тригонометриро ҳал кунед:
\(\cos 3x-\sin x=\sqrt{3}(\cos x-\sin 3x)\)
Ҳал.
\(\cos 3x-\sin x=\sqrt{3}\cos x-\sqrt{3}\sin 3x\)
\(\cos 3x+\sqrt{3}\sin 3x=\sqrt{3}\cos x+\sin x \quad |\cdot\frac{1}{2}\)
\(\frac{1}{2}\cdot(\cos 3x+\sqrt{3}\sin 3x)=\frac{1}{2}\cdot(\sqrt{3}\cos x+\)
\(\quad\quad+\sin x)\)
\(\frac{1}{2}\cos 3x+\frac{1}{2}\sqrt{3}\sin 3x=\frac{1}{2}\sqrt{3}\cos x+\frac{1}{2}\sin x\)
\(\sin\frac{\pi}{6}\cos 3x+\cos\frac{\pi}{6}\sin 3x=\sin\frac{\pi}{3}\cos x+\)
\(\quad\quad+\cos\frac{\pi}{3}\sin x\)
\(\sin(\frac{\pi}{6}+3x)=\sin(\frac{\pi}{3}+x)\)
\(\sin(\frac{\pi}{6}+3x)-\sin(\frac{\pi}{3}+x)=0\)
\(2\sin\frac{\frac{\pi}{6}+3x-(\frac{\pi}{3}+x)}{2}\cos\frac{\frac{\pi}{6}+3x+\frac{\pi}{3}+x}{2}=0\)
\(2\sin\frac{\frac{\pi}{6}+3x-\frac{\pi}{3}-x}{2}\cos\frac{\frac{\pi}{6}+\frac{\pi}{3}+4x}{2}=0\)
\(2\sin\frac{\frac{\pi}{6}\frac{\pi}{3}+2x}{2}\cos\frac{\frac{\pi}{6}+\frac{\pi}{3}+4x}{2}=0\)
1)\(\frac{\pi}{6}-\frac{\pi}{3}=\frac{\pi}{6}-\frac{2\pi}{6}=\frac{\pi-2\pi}{6}=\frac{-\pi}{6}=-\frac{\pi}{6}\);
1)\(\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{6}+\frac{2\pi}{6}=\frac{\pi+2\pi}{6}=\frac{3\pi}{6}=-\frac{\pi}{2}\).
\(2\sin\frac{-\frac{\pi}{6}+2x}{2}\cos\frac{\frac{\pi}{2}+4x}{2}=0\)
\(2\sin(\frac{-\frac{\pi}{6}}{2}+\frac{2x}{2})\cos(\frac{\frac{\pi}{2}}{2}+\frac{4x}{2})=0\)
\(2\sin(-\frac{\pi}{12}+x)\cos(\frac{\pi}{4}+2x)=0\quad |\cdot\frac{1}{2}\)
\(\sin(-\frac{\pi}{12}+x)\cos(\frac{\pi}{4}+2x)=0\)
Ҳолати якумро дида мебароем:
1) \(\sin(-\frac{\pi}{12}+x)=0\)
\(-\frac{\pi}{12}+x=\pi n\)
\(x=\pi n+\frac{\pi}{12}\)
\(x=\frac{\pi}{12}(12n+1)\).
Акнун ҳолати дуюмро низ дида мебароем:
2) \(\cos(\frac{\pi}{4}+2x)=0\)
\(\frac{\pi}{4}+2x=\frac{\pi}{2}+\pi n\)
\(2x=\frac{\pi}{2}+\pi n-\frac{\pi}{4}\)
\(2x=\frac{2\pi}{4}-\frac{\pi}{4}+\pi n\)
\(2x=\frac{2\pi-\pi}{4}+\pi n\)
\(2x=\frac{\pi}{4}+\pi n\quad |\cdot\frac{1}{2}\)
\(\frac{1}{2}\cdot2x=\frac{1}{2}\cdot(\frac{\pi}{4}+\pi n)\)
\(x=\frac{1}{2}\cdot\frac{\pi}{4}+\frac{1}{2}\cdot\pi n\)
\(x=\frac{\pi}{8}+\frac{\pi n}{2}\)
\(x=\frac{\pi n}{2}+\frac{\pi}{8}\)
\(x=\frac{\pi}{8}(4n+1)\).
Ҷавоб: \(x=\frac{\pi}{12}(12n+1)\);\(x=\frac{\pi}{8}(4n+1)\), \(n\in \mathbb{Z}\).