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cos6x . cos2x + \(\dfrac{1}{2}\) = 0
⇔ 2cos6x . cos2x + 1 = 0
⇔ cos8x + cos4x + 1 = 0
⇔ 2cos24x + cos4x = 0
⇔ \(\left[{}\begin{matrix}cos4x=0\\cos4x=-\dfrac{1}{2}\end{matrix}\right.\)
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c/
\(\Leftrightarrow sinx+sin3x+sin2x=cosx+cos3x+cos2x\)
\(\Leftrightarrow2sin2x.cosx+sin2x=2cos2x.cosx+cos2x\)
\(\Leftrightarrow sin2x\left(2cosx+1\right)=cos2x\left(2cosx+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx+1=0\\sin2x=cos2x=sin\left(\frac{\pi}{2}-2x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\2x=\frac{\pi}{2}-2x+k2\pi\\2x=2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{2\pi}{3}+k2\pi\\x=\frac{\pi}{8}+\frac{k\pi}{2}\\\end{matrix}\right.\)
b/
\(\Leftrightarrow sin2x+sin6x-\left(cos5x+cosx\right)=0\)
\(\Leftrightarrow2sin4x.cos2x-2cos3x.cos2x=0\)
\(\Leftrightarrow cos2x\left(sin4x-cos3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin4x=cos3x=sin\left(\frac{\pi}{2}-3x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\4x=\frac{\pi}{2}-3x+k2\pi\\4x=3x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{14}+\frac{k2\pi}{7}\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
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\(pt\Leftrightarrow cos6x+3cos2x-4\left(2cos^2x-1\right)=0\)
\(\Leftrightarrow cos6x+3cos2x-4cos2x=0\)
\(\Leftrightarrow cos6x-cos2x=0\)
\(\Leftrightarrow-2sin4x.sin2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{k\pi}{4}\)
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c/
\(\Leftrightarrow\sqrt{3}sin3x-cos3x=sin2x-\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin3x-\frac{1}{2}cos3x=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)=sin\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=2x-\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{6}=\pi-2x+\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
e/
\(\Leftrightarrow\frac{1}{2}sin8x-\frac{\sqrt{3}}{2}cos8x=\frac{\sqrt{3}}{2}sin6x+\frac{1}{2}cos6x\)
\(\Leftrightarrow sin\left(8x-\frac{\pi}{3}\right)=sin\left(6x+\frac{\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}8x-\frac{\pi}{3}=6x+\frac{\pi}{6}+k2\pi\\8x-\frac{\pi}{3}=\pi-6x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{28}+\frac{k\pi}{7}\end{matrix}\right.\)
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\(1+sinx-cos2x=0\)
\(\Leftrightarrow1+sinx-\left(1-2sin^2x\right)=0\)
\(\Leftrightarrow sinx\left(1+2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(sin3x-sinx+cos2x=0\)
\(\Leftrightarrow2cos2x.sinx+cos2x=0\)
\(\Leftrightarrow cos2x\left(2sinx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sinx=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
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Phương trình đã cho tương đương với:
\(cos2x+\left(cos6x+cos10x\right)=0\)
\(\Leftrightarrow cos2x+2.cos8x.cos2x=0\)
\(\Leftrightarrow cos2x\left(1+2cos8x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\1+2cos8x=0\end{matrix}\right.\)
+ TH1:
\(cos2x=0\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\left(k\in Z\right)\)
+ TH2:
\(1+2cos8x=0\Leftrightarrow cos8x=-\dfrac{1}{2}=cos\dfrac{2\pi}{3}\)
\(\Leftrightarrow8x=\pm\dfrac{2\pi}{3}+k2\pi\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+\dfrac{k\pi}{4}\\x=-\dfrac{\pi}{12}+\dfrac{k\pi}{4}\end{matrix}\right.\) \(\left(k\in Z\right)\)
Vậy phương trình gồm các họ nghiệm: \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\), \(x=\dfrac{\pi}{12}+\dfrac{k\pi}{4}\), \(x=-\dfrac{\pi}{12}+\dfrac{k\pi}{4}\) với \(k\in Z\)
ĐKXĐ: ....
\(\Leftrightarrow\frac{cos2x}{sin3x}+\frac{cos2\left(3x\right)}{sin3\left(3x\right)}+\frac{cos2\left(9x\right)}{sin3\left(9x\right)}=0\)
Xét biểu thức \(\frac{cos2a}{sin3a}=\frac{cos2a.sina}{sin3a.sina}=\frac{sin3a-sina}{2sin3a.sina}=\frac{1}{2}\left(\frac{1}{sina}-\frac{1}{sin3a}\right)\)
Vậy pt tương đương:
\(\frac{1}{2}\left(\frac{1}{sinx}-\frac{1}{sin3x}+\frac{1}{sin3x}-\frac{1}{sin9x}+\frac{1}{sin9x}-\frac{1}{sin27x}\right)=0\)
\(\Leftrightarrow\frac{1}{sinx}=\frac{1}{sin27x}\Leftrightarrow sinx=sin27x\Leftrightarrow...\)