\(pt\Leftrightarrow cos6x+3cos2x-4\left(2cos^2x-1\right)=0\)

\(\Leftrightarrow cos6x+3cos2x-4cos2x=0\)

\(\Leftrightarrow cos6x-cos2x=0\)

\(\Leftrightarrow-2sin4x.sin2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin2x=0\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{k\pi}{4}\)

\( 2)\sin x + \sin 2x + \sin 3x = 0\\ \Leftrightarrow 2\sin 2x.\cos x + \sin 2x = 0\\ \Leftrightarrow \sin 2x\left( {2\cos x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin 2x = 0\\ 2\cos x + 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 2x = k\pi \\ \cos x = \dfrac{{ - 1}}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{k\pi }}{2}\\ x = \pm \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z} } \right) \)

\( 3)\sin x + \sin 2x + \sin 3x + \sin 4x = 0\\ \Leftrightarrow \left( {\sin x + \sin 4x} \right) + \left( {\sin 2x + \sin 3x} \right) = 0\\ \Leftrightarrow 2\sin \dfrac{{5x}}{2}.\cos \dfrac{{3x}}{2} + 2\sin \dfrac{{5x}}{2}.\cos \dfrac{x}{2} = 0\\ \Leftrightarrow \sin \dfrac{{5x}}{2}.\left( {\cos \dfrac{{3x}}{2} + \cos \dfrac{x}{2}} \right) = 0\\ \Leftrightarrow \sin \dfrac{{5x}}{2}.2\cos x.\cos \dfrac{x}{2} = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin \dfrac{{5x}}{2} = 0\\ 2\cos x = 0\\ \cos \dfrac{x}{2} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{2k\pi }}{5}\\ x = \dfrac{\pi }{2} + k\pi \\ x = \pi + 2k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \)

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:

\(\frac{1}{2}\left(\frac{1}{sinx}-\frac{1}{sin3x}+\frac{1}{sin3x}-\frac{1}{sin9x}+\frac{1}{sin9x}-\frac{1}{sin27x}\right)=0\)

\(\Leftrightarrow\frac{1}{sinx}=\frac{1}{sin27x}\Leftrightarrow sinx=sin27x\Leftrightarrow...\)

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 \(k\in Z\)

1.

\(\sin ^8x-\cos ^8x=(\sin ^4x+\cos ^4x)(\sin ^4x-\cos ^4x)\)

\(=[(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x](\sin ^2x+\cos ^2x)(\sin ^2x-\cos ^2x)\)

\(=(1-2\sin ^2x\cos ^2x)(\sin ^2x-\cos ^2x)\)

\(=(1-\frac{\sin ^22x}{2})(-\cos 2x)=-\frac{(2-\sin ^22x)\cos 2x}{2}=-\frac{(1+\cos ^22x)\cos 2x}{2}\) (1)

\(-(\frac{7}{8}\cos 2x+\frac{1}{8}\cos 6x)=\frac{-7}{8}\cos 2x-\frac{1}{8}(4\cos ^32x-3\cos 2x)=-\frac{\cos 2x+\cos ^32x}{2}\)

\(=\frac{-\cos 2x(\cos ^22x+1)}{2}\) (2)

Từ $(1);(2)$ ta có đpcm.

2.

\(\text{VP}=\frac{1}{32}(2+\cos 2x-2\cos 4x-\cos 6x)\)

\(=\frac{1}{32}[2+\cos 2x-2(2\cos ^22x-1)-(4\cos ^32x-3\cos 2x)]\)

\(=\frac{1}{8}(-\cos ^32x-\cos ^22x+\cos 2x+1)=\frac{1}{8}(\cos 2x+1)(1-\cos ^22x)=\frac{1}{8}(\cos 2x+1)\sin ^22x\) (1)

\(\text{VT}=\sin ^2x\cos ^4x=\frac{1}{8}.(2\sin x\cos x)^2.2\cos ^2x=\frac{1}{8}\sin ^22x.(\cos 2x+1)(2)\)

Từ $(1);(2)$ ta có đpcm.

ĐKXĐ: \(x\ne k\pi\)

\(sin7x=sin^2x+2sinx.cos2x+2sinx.cos4x+2sinx.cos6x\)

\(\Leftrightarrow sin7x=sin^2x+sin3x-sinx+sin5x-sin3x+sin7x-sin5x\)

\(\Leftrightarrow sin7x=sin^2x-sinx+sin7x\)

\(\Leftrightarrow sinx\left(sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(loại\right)\\sinx=1\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)