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Y
2
HP
24 tháng 8 2021
1.
\(8sinx=\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}\)
\(\Leftrightarrow4sinx=\dfrac{\sqrt{3}}{2cosx}+\dfrac{1}{2sinx}\)
\(\Leftrightarrow4sinx=\dfrac{\sqrt{3}sinx+cosx}{sin2x}\)
\(\Leftrightarrow4sinx.sin2x=\sqrt{3}sinx+cosx\)
\(\Leftrightarrow2cosx-2cos3x=\sqrt{3}sinx+cosx\)
\(\Leftrightarrow cosx-\sqrt{3}sinx=2cos3x\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos3x\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=\pm3x+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}-k\pi\\x=-\dfrac{\pi}{12}+\dfrac{k\pi}{2}\end{matrix}\right.\)
HP
24 tháng 8 2021
2.
ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(sinx+\sqrt{3}cosx=\dfrac{1}{cosx}\)
\(\Leftrightarrow2sinx.cosx+2\sqrt{3}cos^2x-\sqrt{3}=2-\sqrt{3}\)
\(\Leftrightarrow\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x=1-\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac{2-\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k2\pi\\2x+\dfrac{\pi}{3}=\pi-arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+\dfrac{1}{2}arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k\pi\\x=\dfrac{\pi}{3}-\dfrac{1}{2}arcsin\left(\dfrac{2-\sqrt{3}}{2}\right)+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
2.1
a.
\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
b.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
HP
5 tháng 9 2021
1.
\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)=1-sinx.cosx\)
\(\Leftrightarrow\left(1-sinx.cosx\right)\left(sinx+cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx.cosx=1\\sinx+cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
HP
5 tháng 9 2021
2.
\(\left|cosx-sinx\right|+2sin2x=1\)
\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)
\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)
\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=k\pi\\cos^2x+sin^2x-2sinx.cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)
I
21 tháng 9 2023
a,
\(\cos^3x-\sin^3x=\cos x+\sin x\\ < =>\cos^3x-\cos x=\sin^3x-\sin x\\ < =>\cos x\left(\cos^2x-1\right)=\sin x\left(\sin^2x-1\right)\\ < =>\cos x.\left(-\sin^2x\right)=\sin x.\left(-\cos^2x\right)\\ < =>\dfrac{1}{cosx}=\dfrac{1}{sinx}\)
b,
\(2sinx+2\sqrt{3}cosx=\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}\\ < =>2sinx-\dfrac{1}{sinx}=\dfrac{\sqrt{3}}{cosx}-2\sqrt{3}cosx\\ < =>\dfrac{2sin^2x-1}{sinx}=\dfrac{\sqrt{3}.cosx.\left(1-2cos^2x\right)}{cosx}\\ < =>\dfrac{cos2x}{sinx}=\sqrt{3}.cos2x\\ < =>\dfrac{1}{sinx}=\sqrt{3}\)
Đặt \(\sqrt{3}\sin x+\cos x=a\)
Theo đề, ta có: \(a=3+\dfrac{1}{a+1}=\dfrac{3a+3+1}{a+1}=\dfrac{3a+4}{a+1}\)
\(\Leftrightarrow a^2+a-3a-4=0\)
\(\Leftrightarrow a^2-2a-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1+\sqrt{5}\\a=1-\sqrt{5}\end{matrix}\right.\)
TH1: \(a=1+\sqrt{5}\)
\(\Leftrightarrow\sqrt{3}\sin x+\cos x=\sqrt{5}+1\)(1)
Vì \(3+1=4< 6+2\sqrt{5}\)
nên (1) vô nghiệm
TH2: \(a=1-\sqrt{5}\)
\(\Leftrightarrow\sqrt{3}\sin x+1\cos x=1-\sqrt{5}\)
\(\Leftrightarrow\sin\left(x+\dfrac{\Pi}{6}\right)=\dfrac{1-\sqrt{5}}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{\Pi}{6}=arcsin\left(\dfrac{1-\sqrt{5}}{2}\right)+k2\Pi\\x+\dfrac{\Pi}{6}=\Pi-arcsin\left(\dfrac{1-\sqrt{5}}{2}\right)+k2\Pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=arcsin\left(\dfrac{1-\sqrt{5}}{2}\right)+k2\Pi-\dfrac{\Pi}{6}\\x=-arcsin\left(\dfrac{1-\sqrt{5}}{2}\right)+\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)