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Муодилаи тригонометриро ҳал кунед:
\(\frac{4\operatorname{ctg}x}{1+\operatorname{ctg}^{2}x}+\sin^{2}2x+1=0\).
Ҳал.
1) \(\frac{4\operatorname{ctg}x}{1+\operatorname{ctg}^{2}x}= \frac{4\operatorname{ctg}x}{\frac{1}{\sin^{2}x}}=4\operatorname{ctg}x:\frac{1}{\sin^{2}x}=\)
\(=4\operatorname{ctg}x\cdot\sin^{2}x= 4\cdot\frac{\cos x}{\sin x}\cdot\sin^{2}x=\)
\(=4\cos x\sin x = 4\sin x\cos x\);
2) \(\sin^{2}2x=(\sin 2x)^2=(2\sin x\cos x)^2=\)
\(=4\cdot(\sin x\cos x)^2\).
Аз ин ду баробарӣ мебарояд, ки
\(\frac{4\operatorname{ctg}x}{1+\operatorname{ctg}^{2}x}+\sin^{2}2x+1=\)
\(=4\sin x\cos x+4\cdot(\sin x\cos x)^2+1=\)
\(=4\cdot(\sin x\cos x)^2+4\sin x\cos x+1\).
\(4\cdot(\sin x\cos x)^2+4\sin x\cos x+1=0\).
Бигзор \(t = \sin x\cos x\). Пас,
\(4t^2+4t+1=0\)
\(D=b^2-4\cdot a\cdot c=4^2-4\cdot4\cdot1=\)
\(\quad=16-16=0\), \(D=0\)
\(t=\frac{-b}{2\cdot a}=\frac{-4}{2\cdot4}=\frac{-4}{8}=\frac{-1}{2}=-\frac{1}{2}\).

\(\sin x\cos x=-\frac{1}{2}\quad |\cdot2\)
\(2\sin x\cos x=-\frac{1}{2}\cdot2\)
\(2\sin x\cos x=-1\)
\(\sin 2x=-1\)
\(2x=-\frac{\pi}{2}+2\pi n\quad |\cdot\frac{1}{2}\)
\(\frac{1}{2}\cdot2x=\frac{1}{2}\cdot(-\frac{\pi}{2}+2\pi n)\)
\(x=\frac{1}{2}\cdot(-\frac{\pi}{2})+\frac{1}{2}\cdot2\pi n\)
\(x=-\frac{\pi}{4}+\frac{2\pi n}{2}\)
\(x=-\frac{\pi}{4}+\pi n\)
\(x=\pi n-\frac{\pi}{4}\)
\(x=\frac{\pi}{4}(4n-1)\).
Ҷавоб: \(x=\frac{\pi}{4}(4n-1)\), \(n\in \mathbb{Z}\).