a/

\(\Leftrightarrow1+cos2x+cos3x+cosx=0\)

\(\Leftrightarrow2cos^2x+2cos2x.cosx=0\)

\(\Leftrightarrow2cosx\left(cosx+cos2x\right)=0\)

\(\Leftrightarrow2cosx\left(2cos^2x+cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=-1\\cosx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)

b/

\(\Leftrightarrow2sin3x.cosx+sin3x=2cos3x.cosx+cos3x\)

\(\Leftrightarrow sin3x\left(2cosx+1\right)-cos3x\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left(sin3x-cos3x\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(3x-\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)

c/

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x=1-cos4x\)

\(\Leftrightarrow cos6x+cos2x-2cos4x=0\)

\(\Leftrightarrow2cos4x.cos2x-2cos4x=0\)

\(\Leftrightarrow2cos4x\left(cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)

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

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

\(cosx+cos3x+cos2x+cos4x=0\)

\(\Leftrightarrow2cos2x.cosx+2cos3x.cosx=0\)

\(\Leftrightarrow cosx.\left(cos2x+cos3x\right)=0\)

\(\Leftrightarrow cosx.cos\frac{5x}{2}.cos\frac{x}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos\frac{5x}{2}=0\\cos\frac{x}{2}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\\x=\pi+k2\pi\end{matrix}\right.\)

\(sinx+sin7x+sin3x+sin5x=0\)

\(\Leftrightarrow2sin4x.cos3x+2sin4x.cosx=0\)

\(\Leftrightarrow sin4x\left(cos3x+cosx\right)=0\)

\(\Leftrightarrow sin4x.cos2x.cosx=0\)

\(\Leftrightarrow sin4x=0\)

\(\Rightarrow4x=k\pi\Rightarrow x=\frac{k\pi}{4}\)

Lý do chỉ cần 1 pt sin4x=0 do sin4x bao hàm cả cosx và cos2x ở trong đó

2.

\(sin3x+cos2x=1+2sinx.cos2x\)

\(\Leftrightarrow sin3x+cos2x=1+sin3x-sinx\)

\(\Leftrightarrow cos2x+sinx-1=0\)

\(\Leftrightarrow-2sin^2x+sinx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

1.

\(cos3x-cos4x+cos5x=0\)

\(\Leftrightarrow cos3x+cos5x-cos4x=0\)

\(\Leftrightarrow2cos4x.cosx-cos4x=0\)

\(\Leftrightarrow\left(2cosx-1\right)cos4x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\cos4x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\4x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\end{matrix}\right.\)

3.

\(cos2x-cosx=2sin^2\dfrac{3x}{2}\)

\(\Leftrightarrow2sin\dfrac{3x}{2}.sin\dfrac{x}{2}+2sin^2\dfrac{3x}{2}=0\)

\(\Leftrightarrow2sin\dfrac{3x}{2}.\left(sin\dfrac{x}{2}+sin\dfrac{3x}{2}\right)=0\)

\(\Leftrightarrow sin\dfrac{3x}{2}.sinx.cos\dfrac{x}{2}=0\)

Đến đây dễ rồi tự làm tiếp nha.

a/

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin5x-\frac{1}{2}cos5x=-1\)

\(\Leftrightarrow sin\left(5x-\frac{\pi}{6}\right)=-1\)

\(\Leftrightarrow5x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=-\frac{\pi}{15}+\frac{k2\pi}{5}\)

b/

\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=\frac{1}{2}\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{\sqrt{2}}\)

\(\Leftrightarrow sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{4}=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{7\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{36}+\frac{k2\pi}{3}\end{matrix}\right.\)

d/

\(\Leftrightarrow2sinx.cosx+1-2sin^2x=1\)

\(\Leftrightarrow2sinx\left(cosx-sinx\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)

1.

\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)

\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}cos3x+\sqrt{2}sin3x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{\sqrt{2}}cos3x+\dfrac{1}{\sqrt{2}}sin3x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin\left(3x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=3x+\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{3}=\pi-3x-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)

2.

\(sinx-\sqrt{3}cosx=2sin5\text{​​}x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=5x+k2\pi\\x-\dfrac{\pi}{3}=\pi-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)

\(D=\frac{sin4x+sin5x+sin6x}{cos4x+cos5x+cos6x}\)

\(=\frac{\left(sin4x+sin6x\right)+sin5x}{\left(cos4x+cos6x\right)+cos5x}\)

\(=\frac{2sin\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+sin5x}{2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}+cos5x}\)

\(=\frac{2sin5x.cos\left(-x\right)+sin5x}{2cos5x.cos\left(-x\right)+cos5x}=\frac{sin5x\left(2.cos\left(-x\right)+1\right)}{cos5x\left(2.cos\left(-x\right)+1\right)}=\frac{sin5x}{cos5x}=tan5x\)