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LC
27 tháng 9 2018
3.3 d)
\(\sin8x-\cos6x=\sqrt{3}\left(\sin6x+\cos8x\right)\\ \Leftrightarrow\sin8x-\sqrt{3}\cos8x=\sqrt{3}\sin6x+\cos6x\\ \Leftrightarrow\sin\left(8x-\dfrac{\pi}{3}\right)=\sin\left(6x+\dfrac{\pi}{6}\right)\\ \Leftrightarrow\left[{}\begin{matrix}8x-\dfrac{\pi}{3}=6x+\dfrac{\pi}{6}+k2\pi\\8x-\dfrac{\pi}{3}=\pi-\left(6x+\dfrac{\pi}{6}\right)+k2\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k\dfrac{\pi}{7}\end{matrix}\right.\)
LC
27 tháng 9 2018
3.4 a)
\(2sin\left(x+\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(\dfrac{\pi}{2}-x-\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(-x+\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \Leftrightarrow2cos\left(x-\dfrac{\pi}{4}\right)+4sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3\sqrt{2}}{5}\\ \)
Chia hai vế cho \(\sqrt{2^2+4^2}=2\sqrt{5}\)
Ta được:
\(\dfrac{1}{\sqrt{5}}cos\left(x-\dfrac{\pi}{4}\right)+\dfrac{2}{\sqrt{5}}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{3}{4}\\ \)
Gọi \(\alpha\) là góc có \(cos\alpha=\dfrac{1}{\sqrt{5}}\)và \(sin\alpha=\dfrac{2}{\sqrt{5}}\)
Phương trình tương đương:
\(cos\left(x-\dfrac{\pi}{4}-\alpha\right)=\dfrac{3}{4}\\ \Leftrightarrow x=\pm arscos\left(\dfrac{3}{4}\right)+\dfrac{\pi}{4}+\alpha+k2\pi\)
T
11 tháng 7 2018
\(1+\sin\dfrac{x}{2}\sin x-\cos\dfrac{x}{2}\sin^2x=2\cos^2\left(\dfrac{\Pi}{4}-\dfrac{x}{2}\right)\)
\(\Leftrightarrow1+\sin\dfrac{x}{2}\sin x-\cos\dfrac{x}{2}\sin^2x=2\left(\dfrac{\sqrt{2}}{2}\cos\dfrac{x}{2}+\dfrac{\sqrt{2}}{2}\sin\dfrac{x}{2}\right)^2\)
\(\Leftrightarrow1+2\sin^2\dfrac{x}{2}\cos\dfrac{x}{2}-\cos\dfrac{x}{2}\left(2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\right)^2=1+2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\)
\(\Leftrightarrow2\sin^2\dfrac{x}{2}\cos\dfrac{x}{2}-4\cos^3\dfrac{x}{2}\sin^2\dfrac{x}{2}-2\sin\dfrac{x}{2}\cos\dfrac{x}{2}=0\)
\(\Leftrightarrow2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\left(\sin\dfrac{x}{2}-2\sin\dfrac{x}{2}\cos^2\dfrac{x}{2}-1\right)=0\)
\(\Leftrightarrow2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\left(\sin\dfrac{x}{2}-2\sin\dfrac{x}{2}\left(1-\sin^2\dfrac{x}{2}\right)-1\right)=0\)
\(\Leftrightarrow2\sin\dfrac{x}{2}\cos\dfrac{x}{2}.\left(\sin\dfrac{x}{2}-1\right)\left(2\sin^2\dfrac{x}{2}+2\sin\dfrac{x}{2}+1\right)=0\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2020
c.
\(\Leftrightarrow tanx=-\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow x=-\frac{\pi}{6}+k\pi\)
d.
\(\Leftrightarrow\frac{1}{2}sin2x.cos2x=0\)
\(\Leftrightarrow\frac{1}{4}sin4x=0\)
\(\Leftrightarrow sin4x=0\)
\(\Leftrightarrow x=\frac{k\pi}{4}\)
e.
\(\Leftrightarrow4sin4x.cos4x=-\sqrt{2}\)
\(\Leftrightarrow2sin8x=-\sqrt{2}\)
\(\Leftrightarrow sin8x=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}8x=-\frac{\pi}{4}+k2\pi\\8x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{32}+\frac{k\pi}{4}\\x=\frac{5\pi}{32}+\frac{k\pi}{4}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2020
a.
\(\Leftrightarrow sin2x=cosx\)
\(\Leftrightarrow sin2x=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-x+k2\pi\\2x=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
b.
\(\Leftrightarrow\left[{}\begin{matrix}cos\frac{x}{2}=\frac{1}{2}\\sin\frac{x}{2}=-2< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{2}=\frac{\pi}{3}+k2\pi\\\frac{x}{2}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k4\pi\\x=-\frac{2\pi}{3}+k4\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
c.
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=-sin\left(x-\frac{2\pi}{5}-\pi\right)\)
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=sin\left(x-\frac{2\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{2\pi}{3}=x-\frac{2\pi}{5}+k2\pi\\3x+\frac{2\pi}{3}=\frac{7\pi}{5}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{8\pi}{15}+k\pi\\x=\frac{11\pi}{60}+\frac{k\pi}{2}\end{matrix}\right.\)
d.
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=\frac{\pi}{4}+x+k2\pi\\4x+\frac{\pi}{3}=-\frac{\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2020
a.
\(sin\left(2x+1\right)=-cos\left(3x-1\right)\)
\(\Leftrightarrow sin\left(2x+1\right)=sin\left(3x-1-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1-\frac{\pi}{2}=2x+1+k2\pi\\3x-1-\frac{\pi}{2}=\pi-2x-1+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+2+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
b.
\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
21 tháng 7 2022
b: \(\Leftrightarrow\dfrac{1-\cos2x}{2}+\dfrac{1-\cos4x}{2}+\dfrac{1-\cos6x}{2}=\dfrac{3}{2}\)
\(\Leftrightarrow3-\cos2x-\cos4x-\cos6x=3\)
\(\Leftrightarrow\cos2x+\cos6x+\cos4x=0\)
\(\Leftrightarrow2\cdot\cos\left(\dfrac{6x+2x}{2}\right)\cdot\cos\left(\dfrac{6x-2x}{2}\right)+\cos2x=0\)
\(\Leftrightarrow\cos2x\left(2\cdot\cos4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\\cos4x=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{4}+\dfrac{k\Pi}{2}\\4x=-\dfrac{2}{3}\Pi+k2\Pi\\4x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{4}+\dfrac{k\Pi}{2}\\x=-\dfrac{1}{6}\Pi+\dfrac{k\Pi}{2}\\x=\dfrac{1}{6}\Pi+\dfrac{k\Pi}{2}\end{matrix}\right.\)
\(\sin x=\dfrac{1}{2}\Leftrightarrow\sin x=\sin\dfrac{\pi}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)