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a.
\(cos\left(3x-\frac{\pi}{6}\right)=sin\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(3x-\frac{\pi}{6}\right)=cos\left(\frac{\pi}{6}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=\frac{\pi}{6}-2x+k2\pi\\3x-\frac{\pi}{6}=2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\cos3x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne\frac{1}{2}\end{matrix}\right.\)
\(tan3x-tanx=0\)
\(\Leftrightarrow\frac{sin3x}{cos3x}-\frac{sinx}{cosx}=0\)
\(\Leftrightarrow sin3x.cosx-cos3x.sinx=0\)
\(\Leftrightarrow sin2x=0\)
\(\Leftrightarrow2sinx.cosx=0\)
\(\Leftrightarrow sinx=0\Leftrightarrow x=k\pi\)
c.
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{2\pi}{5}\right)=\frac{1}{2}-\frac{1}{2}cos\left(4x+\frac{8\pi}{5}\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=-cos\left(4x+\frac{3\pi}{5}+\pi\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=cos\left(4x+\frac{3\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{3\pi}{5}=2x-\frac{2\pi}{5}+k2\pi\\4x+\frac{3\pi}{5}=\frac{2\pi}{5}-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
d.
\(\Leftrightarrow cos^2\left(2x-1\right)=0\)
\(\Leftrightarrow cos\left(2x-1\right)=0\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{1}{2}+\frac{k\pi}{2}\)
ĐKXĐ: \(sinx\ne0\)
\(\Leftrightarrow cot^2x=\left|\frac{1-\left|sinx\right|}{1-\left|cosx\right|}\right|\Leftrightarrow cot^2x=\frac{1-\left|sinx\right|}{1-\left|cosx\right|}\)
Đặt \(\left\{{}\begin{matrix}\left|sinx\right|=a\Rightarrow0< a\le1\\\left|cosx\right|=b\Rightarrow0\le b< 1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\frac{b^2}{a^2}=\frac{1-a}{1-b}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\frac{b^2}{1-b^2}=\frac{1-a}{1-b}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\frac{b^2}{\left(1-b\right)\left(1+b\right)}=\frac{\left(1-a\right)\left(1+b\right)}{\left(1-b\right)\left(1+b\right)}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\1-a^2=\left(1-a\right)\left(1-b\right)\end{matrix}\right.\)
\(\Rightarrow\left(1-a\right)\left(1+a\right)=\left(1-a\right)\left(1-b\right)\)
\(\Rightarrow\left[{}\begin{matrix}a=1\\1+a=1-b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left|sinx\right|=1\Leftrightarrow sin^2x=1\Leftrightarrow cosx=0\Leftrightarrow x=...\)
ĐKXĐ: \(x\ne\frac{\pi}{6}+\frac{k\pi}{3}\)
\(\Leftrightarrow\frac{cos^2x-cos3x.cos5x}{cos3x.cosx}-4\left[1-2sin^2\left(2x+\frac{11\pi}{2}\right)\right]-4cos2x=0\)
\(\Leftrightarrow\frac{2cos^2x-cos2x-cos8x}{cos4x+cos2x}-4cos\left(4x+11\pi\right)-4cos2x=0\)
\(\Leftrightarrow\frac{1-cos8x}{cos4x+cos2x}+4cos4x-4cos2x=0\)
\(\Leftrightarrow1-cos8x+4\left(cos4x-cos2x\right)\left(cos4x+cos2x\right)=0\)
\(\Leftrightarrow1-cos8x+4cos^24x-4cos^22x=0\)
\(\Leftrightarrow1-\left(2cos^24x-1\right)+4cos^24x-2\left(1+cos4x\right)=0\)
\(\Leftrightarrow cos^24x-cos4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos4x=1\end{matrix}\right.\) \(\Leftrightarrow...\)
ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne0\\cos3x+cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sinx\ne0\\2cos2x.cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cos2x\ne0\end{matrix}\right.\)
Phương trình tương đương:
\(\left(sin3x-sinx\right)\left(\frac{cos^2x}{sin^2x}-1\right)=0\)
\(\Leftrightarrow2cos2x.sinx.\left(cos^2x-sin^2x\right)=0\)
\(\Leftrightarrow2cos^22x.sinx=0\)
Phương trình vô nghiệm