\(\dfrac{1}{sinx}+\dfrac{1}{sin2x}+\dfrac{1}{sin4x}=0\)

\(\dfrac{1}{sinx}+cotx+\dfrac{1}{sin2x}+cot2x+\dfrac{1}{sin4x}+cot4x=cotx+cot2x+cot4x\)

\(\dfrac{1+cosx}{sinx}+\dfrac{1+cos2x}{sin2x}+\dfrac{1+cos4x}{sin4x}=cotx+cot2x+cot4x\)

\(\dfrac{2cos^2\dfrac{x}{2}}{2sin\dfrac{x}{2}.cos\dfrac{x}{2}}+\dfrac{2cos^2x}{2sinx.cosx}+\dfrac{2cos^22x}{2sin2x.cos2x}=cotx+cot2x+cot4x\)

\(\dfrac{cos\dfrac{x}{2}}{sin\dfrac{x}{2}}+\dfrac{cosx}{sinx}+\dfrac{cos2x}{sin2x}=cotx+cot2x+cot4x\)

\(cot\dfrac{x}{2}+cotx+cot2x=cotx+cot2x+cot4x\)

\(cot\dfrac{x}{2}=cot4x\)

\(\Rightarrow\dfrac{x}{2}=4x+k\text{π}\)

\(\Leftrightarrow x=-\dfrac{k2\text{π}}{7}\)

nguyễn thị hương giang 

mình trình bày chút, giờ mình ms onl

Cộng cả 2 vế với cot8x

\(\dfrac{1}{sin8x}+cot8x=\dfrac{1+cos8x}{sin8x}=\dfrac{2cos^24x}{2sin4x.cos4x}=cot4x\)

Rồi cot4x lại đi với \(\dfrac{1}{sin4x}\) tạo cot2x ư

........... cứ như thế phương trình sẽ trở thành 

\(cot\dfrac{x}{2}=cot8x\)

a/ Bạn coi lại vế trái đề bài, nhìn không hợp lý

b/ \(\Leftrightarrow\frac{1}{2}sin9x-\frac{1}{2}sinx=\frac{1}{2}sin5x-\frac{1}{2}sinx\)

\(\Leftrightarrow sin9x=sin5x\)

\(\Leftrightarrow\left[{}\begin{matrix}9x=5x+k2\pi\\9x=\pi-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{14}+\frac{k\pi}{7}\end{matrix}\right.\)

c/ \(\Leftrightarrow sin2x-cos2x=cosx-sinx\)

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

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

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

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

d/

\(\Leftrightarrow sin2x=sin6x-sin4x\)

\(\Leftrightarrow2sinx.cosx=2cos5x.sinx\)

\(\Leftrightarrow sinx\left(cosx-cos5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos5x=cosx\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{k\pi}{2}\\x=\frac{k\pi}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{k\pi}{3}\end{matrix}\right.\)

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

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

\(\dfrac{1}{sin2k}=\dfrac{sink}{sink.sin2k}=\dfrac{\left(sin2k-k\right)}{sink.sin2k}=\dfrac{sin2k.cosk-cos2k.sink}{sink.sin2k}\)

\(=\dfrac{cosk}{sink}-\dfrac{cos2k}{sin2k}=cotk-cot2k\)

Do đó pt tương đương:

\(cot\dfrac{x}{2}-cotx+cotx-cot2x+...+cot2^{2017}x-cot^{2018}x=0\)

\(\Leftrightarrow cot\dfrac{x}{2}-cot2^{2018}x=0\)

\(\Leftrightarrow\dfrac{x}{2}=2^{2018}x+k\pi\)

\(\Leftrightarrow...\)

@Nguyễn VIệt Lâm giúp em với

1.

\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)

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

\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)

\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)

Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)

\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)

2.

\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)

\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)

\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)

\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

a: \(\Leftrightarrow2\cdot\sin3x\cdot\cos x-2\cos^2x=0\)

\(\Leftrightarrow\cos x\left(\sin3x-\cos x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\\sin3x=\cos x=\sin\left(\dfrac{\Pi}{2}-x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\3x=\dfrac{\Pi}{2}-x+k2\Pi\\3x=\dfrac{\Pi}{2}+x+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{2}\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)

b: \(\Leftrightarrow\sin x+\sin5x+\sin^2x=0\)

\(\Leftrightarrow\sin x=0\)

hay \(x=k\Pi\)

\(sin4x-2cos2x.cosx=0\)

\(\Leftrightarrow2sin2x.cos2x-2cos2x.cosx=0\)

\(\Leftrightarrow cos2x\left(sin2x-cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\\sin2x-cosx=0\left(1\right)\end{matrix}\right.\)

Xét (1)

\(\Leftrightarrow sin2x=cosx=sin\left(\frac{\pi}{2}-x\right)\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\left(cosx+sin2x\right).sin2x=0\)

\(\Leftrightarrow\left(cosx+2sinx.cosx\right).2sinx.cosx=0\)

\(\Leftrightarrow\left(1+2sinx\right)sinx.cos^2x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin2x=0\end{matrix}\right.\)

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

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

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

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

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

\(\Leftrightarrow2cosx.cos\frac{5x}{2}.cos\frac{x}{2}=0\)

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

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

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

\(\Leftrightarrow2sinx+cos3x+sin2x-sin4x-1=0\)

\(\Leftrightarrow2sinx-1+cos3x-2cos3x.sinx=0\)

\(\Leftrightarrow2sinx-1-cos3x\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(1-cos3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cos3x=1\end{matrix}\right.\)

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