Chương 1: PHÉP DỜI HÌNH VÀ PHÉP ĐỒNG DẠNG TRONG MẶT PHẲNG

Julian Edward

giải pt

a) \(sin^2x+2sin^22x+sin^23x-2=0\)

b) \(2cosx.cos\left(x+\frac{\pi}{3}\right)+\sqrt{3}sin2x=1\)

c) \(5\left(1+cosx\right)=2+sin^4x-cos^4x\)

d) \(1+cot2x=\frac{1-cos2x}{sin^22x}\)

Nguyễn Việt Lâm
25 tháng 8 2020 lúc 21:48

a/

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x-2\left(1-sin^22x\right)=0\)

\(\Leftrightarrow1-\frac{1}{2}\left(cos6x+cos2x\right)-2cos^22x=0\)

\(\Leftrightarrow1-cos4x.cos2x-2cos^22x=0\)

\(\Leftrightarrow2cos^22x-1+cos4x.cos2x=0\)

\(\Leftrightarrow cos4x+cos4x.cos2x=0\)

\(\Leftrightarrow cos4x\left(cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

Nguyễn Việt Lâm
25 tháng 8 2020 lúc 21:51

b/

\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}\right)+\sqrt{3}sin2x=1\)

\(\Leftrightarrow cos2x.cos\left(\frac{\pi}{3}\right)-sin2x.sin\left(\frac{\pi}{3}\right)+\frac{1}{2}+\sqrt{3}sin2x=1\)

\(\Leftrightarrow\frac{1}{2}cos2x+\frac{\sqrt{3}}{2}sin2x=\frac{1}{2}\)

\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\2x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=k\pi\end{matrix}\right.\)

Nguyễn Việt Lâm
25 tháng 8 2020 lúc 21:56

c/

\(\Leftrightarrow5+5cosx=2+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)

\(\Leftrightarrow3+5cosx=sin^2x-cos^2x\)

\(\Leftrightarrow3+5cosx=1-cos^2x-cos^2x\)

\(\Leftrightarrow2cos^2x+5cosx+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-2\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

Nguyễn Việt Lâm
25 tháng 8 2020 lúc 22:00

d/

ĐKXĐ: \(sin2x\ne0\) \(\Leftrightarrow2x\ne k\pi\)

\(\Leftrightarrow1+\frac{cos2x}{sin2x}=\frac{1-cos2x}{sin^22x}\)

\(\Leftrightarrow sin^22x+sin2x.cos2x=1-cos2x\)

\(\Leftrightarrow sin^22x-1+sin2x.cos2x+cos2x=0\)

\(\Leftrightarrow-cos^22x+sin2x.cos2x+cos2x=0\)

\(\Leftrightarrow cos2x\left(sin2x-cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin2x-cos2x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=k\pi\left(l\right)\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)


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