Муодилаи тригонометриро ҳал кунед:
\(7+4\sin x\cos x+1,5(\operatorname{tg}x+\operatorname{ctg}x)=0\).
Ҳал.
\(7+4\sin x\cos x+1,5\operatorname{tg}x+1,5\operatorname{ctg}x=0\)
\(7+4\sin x\cos x+1,5\frac{\sin x}{\cos x}+1,5\frac{\cos x}{\sin x}=0\quad|\cdot\sin x\cos x\)
\(\sin x\cos x\cdot(7+4\sin x\cos x+1,5\cdot\frac{\sin x}{\cos x}+1,5\frac{\cos x}{\sin x})=0\)
\(7\sin x\cos x+4\sin^2 x\cos^2 x+1,5\cdot\sin^2 x+1,5\cos^2 x=0\)
\(7\sin x\cos x+4(\sin x\cos x)^2+1,5\cdot(\sin^2 x+\cos^2 x)=0\)
\(7\sin x\cos x+4(\sin x\cos x)^2+1,5=0\)
\(4(\sin x\cos x)^2+7\sin x\cos x+1,5=0\)
Бигзор \(t = \sin x\cos x\). Пас,
\(4t^2+7t+1,5=0\quad|\cdot2\)
\(8t^2+14t+3=0\)
\(D=b^2-4\cdot a\cdot c=14^2-4\cdot 8\cdot 3=\)
\(\quad=196-96=100=10^2\), \(D>0\)
\(x_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-14\pm \sqrt{10^2}}{2\cdot8}=\frac{-14\pm 10}{16}\)
\(x_{1}=\frac{-14-10}{16}=\frac{-24}{16}=\frac{-3}{2}=-\frac{3}{2}\);
\(x_{2}=\frac{-14+10}{16}=\frac{-4}{16}=\frac{-1}{4}=-\frac{1}{4}\).
Якум ҳолатро дида мебароем:
\(\sin x\cos x=-\frac{3}{2}\quad|\cdot2\)
\(2\sin x\cos x=-3\)
\(\sin 2x=-3\), лекин \(|\sin x|\leq1\).
\(|-3|=3\) ва \(3>1\). Яъне \(\sin 2x\) ба \(-3\) баробар шуда наметавонад.
Ҳолати дуюмро дида мебароем:
\(\sin x\cos x=-\frac{1}{4}\quad|\cdot2\)
\(2\sin x\cos x=-\frac{1}{2}\)
\(\sin 2x=-\frac{1}{2}\)
\(2x=(-1)^n\operatorname{arcsin}(-\frac{1}{2})+\pi n\)
\(2x=(-1)^n\cdot(-\frac{\pi}{6})+\pi n\)
\(2x=(-1)^n\cdot(-1)\cdot\frac{\pi}{6}+\pi n\)
\(2x=(-1)^{n+1}\cdot\frac{\pi}{6}+\pi n\quad|\cdot\frac{1}{2}\)
\(x=(-1)^{n+1}\cdot\frac{\pi}{12}+\frac{\pi n}{2}\)
Ҷавоб: \(x=(-1)^{n+1}\cdot\frac{\pi}{12}+\frac{\pi n}{2}\).