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

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

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

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

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

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

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

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

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

\(\Leftrightarrow sin4x=0\)

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

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

\(\frac{1+sin4x+cos4x}{1-sin4x+cos4x}=\frac{1+2sin2x.cos2x+2cos^22x-1}{1-2sin2x.cos2x+2cos^22x-1}\)

\(=\frac{2cos2x\left(sin2x+cos2x\right)}{2cos2x\left(cos2x-sin2x\right)}=\frac{sin2x+cos2x}{cos2x-sin2x}\)

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

\(\left(sin5x-cos5x\right)^2-\left(sin3x+cos3x\right)^2\)

\(=\left(\sqrt{2}sin\left(5x-\frac{\pi}{4}\right)\right)^2-\left(\sqrt{2}sin\left(3x+\frac{\pi}{4}\right)\right)^2\)

\(=2sin^2\left(5x-\frac{\pi}{4}\right)-2sin^2\left(3x+\frac{\pi}{4}\right)\)

\(=1-cos\left(10x-\frac{\pi}{2}\right)-1+cos\left(6x+\frac{\pi}{2}\right)\)

\(=-sin10x-sin6x=-2sin8x.cos2x\)

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/

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

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

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

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

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

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.

\(\Leftrightarrow cos^2\left(2x-1\right)=0\)

\(\Leftrightarrow cos\left(2x-1\right)=0\)

\(\Leftrightarrow x=\frac{\pi}{4}+\frac{1}{2}+\frac{k\pi}{2}\)

e.

\(cos3x+cosx+cos2x=0\)

\(\Leftrightarrow2cos2x.cosx+cos2x=0\)

\(\Leftrightarrow cos2x\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

f.

\(\Leftrightarrow4sin4x.cos4x=\sqrt{2}\)

\(\Leftrightarrow2sin8x=\sqrt{2}\)

\(\Leftrightarrow sin8x=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow...\)

\(cosx.cos\left(\frac{\pi}{3}-x\right)cos\left(\frac{\pi}{3}+x\right)=\frac{1}{2}cosx\left(cos\frac{2\pi}{3}+cos2x\right)=-\frac{1}{4}cosx+\frac{1}{2}cosx.cos2x\)

\(=-\frac{1}{4}cosx+\frac{1}{4}\left(cos3x+cosx\right)=\frac{1}{4}cos3x\)

\(sin5x-2sinx\left(cos4x+cos2x\right)=sinx.cos4x+cosx.sin4x-2sinx.cos4x-2sinx.cos2x\)

\(=sin4x.cosx-cos4x.sinx-2sinx.cos2x=sin3x-2sinx.cos2x\)

\(=sinx.cos2x+cosx.sin2x-2sinx.cos2x\)

\(=sin2x.cosx-cos2x.sinx=sinx\)

1.

\(\Leftrightarrow sin5x+\sqrt{3}cos5x=-2sin15x\)

\(\Leftrightarrow\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x=-sin15x\)

\(\Leftrightarrow sin\left(5x+\frac{\pi}{3}\right)=sin\left(-15x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+\frac{\pi}{3}=-15x+k2\pi\\5x+\frac{\pi}{3}=\pi+15x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{60}+\frac{k\pi}{10}\\x=-\frac{\pi}{15}+\frac{k\pi}{5}\end{matrix}\right.\)

2.

\(\Leftrightarrow\left(\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x\right)+\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=2\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+sin\left(x+\frac{\pi}{6}\right)=2\)

Do \(\left\{{}\begin{matrix}sin\left(2x-\frac{\pi}{6}\right)\le1\\sin\left(x+\frac{\pi}{6}\right)\le1\end{matrix}\right.\) với mọi x

\(\Rightarrow sin\left(2x-\frac{\pi}{6}\right)+sin\left(x+\frac{\pi}{6}\right)\le2\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}sin\left(2x-\frac{\pi}{6}\right)=1\\sin\left(x+\frac{\pi}{6}\right)=1\end{matrix}\right.\)

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

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

\(\Rightarrow x=\frac{\pi}{3}+k2\pi\)

3.

\(\Leftrightarrow cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin7x+\frac{1}{2}cos7x=\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\)

\(\Leftrightarrow sin\left(7x+\frac{\pi}{6}\right)=sin\left(5x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\7x+\frac{\pi}{6}=\frac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{24}+\frac{k\pi}{6}\end{matrix}\right.\)

a.\(\frac{k\Pi}{2}+\frac{\alpha}{2}\)

b.\(\left\{{}\begin{matrix}x=\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\\x=\Pi-\frac{1}{4}arcsin\left(\frac{1}{3}\right)+\frac{k\Pi}{2}-\frac{1}{8}\end{matrix}\right.\)