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c/
\(\Leftrightarrow cos3x-\sqrt{3}sin3x=\sqrt{3}cos2x-sin2x\)
\(\Leftrightarrow\frac{1}{2}cos3x-\frac{\sqrt{3}}{2}sin3x=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(3x+\frac{\pi}{3}\right)=cos\left(2x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\3x+\frac{\pi}{3}=-2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
b/
\(\Leftrightarrow cosx-\sqrt{3}sinx=sin2x-\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=sin\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{5\pi}{6}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
e/
ĐKXĐ: ...
\(\Leftrightarrow\frac{2sin4x.cos2x}{cos2x}-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow2sin4x-2cos4x=2\sqrt{2}\)
\(\Leftrightarrow sin4x-cos4x=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(4x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(4x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow4x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{3\pi}{16}+\frac{k\pi}{2}\)
d/
Đặt \(sin2x-cos2x=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(\Rightarrow t^2-3t-4=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{3\pi}{4}+k\pi\end{matrix}\right.\)
a.
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(sinx+cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=-1\\2cosx-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)
\(\Leftrightarrow sinx\left(2cosx+1\right)+cosx\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\text{c) }sin3x-\sqrt{3}cos3x=2cos5x\\ \Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=cos5x\\ \Leftrightarrow sin\frac{\pi}{6}\cdot sin3x-cos\frac{\pi}{6}\cdot cos3x=cos5x\\ \Leftrightarrow cos\left(3x+\frac{\pi}{6}\right)=-cos5x\\ \Leftrightarrow cos\left(3x+\frac{\pi}{6}\right)=cos\left(\pi-5x\right)\\ \Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{6}=\pi-5x+m2\pi\\3x+\frac{\pi}{6}=5x-\pi+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{48}+\frac{m\pi}{4}\\x=\frac{7\pi}{12}-n\pi\end{matrix}\right.\)
\(d\text{) }sinx\left(sinx+2cosx\right)=2\\ \Leftrightarrow cos^2x+\left(sinx-cosx\right)^2=0\\ \Leftrightarrow cosx=sinx=0\left(VN\right)\)
\(e\text{) }\sqrt{3}\left(sin2x+cos7x\right)=sin7x-cos2x\\ \Leftrightarrow\sqrt{3}sin2x+cos2x=sin7x-\sqrt{3}cos7x\\ \Leftrightarrow sin2x\cdot\frac{\sqrt{3}}{2}+cos2x\cdot\frac{1}{2}=sin7x\cdot\frac{1}{2}-cos7x\cdot\frac{\sqrt{3}}{2}\\ \Leftrightarrow sin2x\cdot cos\frac{\pi}{3}+cos2x\cdot sin\frac{\pi}{3}=sin7x\cdot cos\frac{\pi}{3}-cos7x\cdot sin\frac{\pi}{3}\\ \Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(7x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=7x-\frac{\pi}{3}+m2\pi\\2x-\frac{\pi}{3}=\frac{4\pi}{3}-7x+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-m2\pi}{5}\\x=\frac{5\pi}{27}+\frac{n2\pi}{9}\end{matrix}\right.\)
\(\text{a) }\sqrt{3}sin2x-cos2x+1=0\\ \Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=-\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos2x-sin\frac{\pi}{3}\cdot sin2x=\frac{1}{2}\\ \Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=cos\frac{\pi}{3}\\ \Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{3}+m2\pi\\2x-\frac{\pi}{3}=-\frac{\pi}{3}+n2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+m\pi\\x=n\pi\end{matrix}\right.\)
\(\text{b) }pt\Leftrightarrow sin4x=\frac{1-4cosx}{3}\\ \Leftrightarrow sin^24x+cos^24x=\left(\frac{1-cos4x}{3}\right)^2+cos^24x=1\\ \Leftrightarrow\left[{}\begin{matrix}cos4x=1\\cos4x=-\frac{4}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos4x=1\\cos4x=-\frac{4}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{arccos\left(-\frac{4}{5}\right)}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
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