\(sin\dfrac{3x}{2}\left(cosx+cos4x+cos7x\right)\)

\(=\)\(sin\dfrac{3x}{2}.cosx+sin\dfrac{3x}{2}.cos4x+sin\dfrac{3x}{2}.cos7x=\dfrac{1}{2}\left[sin\dfrac{x}{2}+sin\dfrac{5x}{2}\right]+\dfrac{1}{2}\left[sin\left(-\dfrac{5x}{2}\right)+sin\dfrac{11x}{2}\right]+\dfrac{1}{2}\left[sin\left(-\dfrac{11x}{2}\right)+sin\dfrac{17x}{2}\right]\)

\(=\dfrac{1}{2}\left(sin\dfrac{x}{2}+sin\dfrac{17x}{2}\right)\)\(=\dfrac{1}{2}.2.sin\dfrac{9x}{2}.cos4x=sin\dfrac{9x}{2}.cos4x\) 

\(sin\dfrac{3x}{2}\left(sinx+sin4x+sin7x\right)\)

\(=sin\dfrac{3x}{2}.sinx+sin\dfrac{3x}{2}.sin4x+sin\dfrac{3x}{2}.sin7x\)\(=\dfrac{1}{2}\left(cos\dfrac{x}{2}-cos\dfrac{5x}{2}\right)+\dfrac{1}{2}\left(cos\dfrac{-5x}{2}-cos\dfrac{11x}{2}\right)+\dfrac{1}{2}\left(cos\dfrac{-11x}{2}-cos\dfrac{17x}{2}\right)\)

\(=\dfrac{1}{2}\left(cos\dfrac{x}{2}-cos\dfrac{17x}{2}\right)\)\(=\dfrac{1}{2}.-2.sin\dfrac{9x}{2}.sin\left(-4x\right)=sin\dfrac{9x}{2}.sin4x\)

Có \(\dfrac{cos7x+cos4x+cosx}{sin7x+sin4x+sinx}\)

\(=\dfrac{sin\dfrac{3x}{2}\left(cos7x+cos4x+cosx\right)}{sin\dfrac{3x}{2}\left(sin7x+sin4x+sinx\right)}\)\(=\dfrac{sin\dfrac{9x}{2}.cos4x}{sin\dfrac{9x}{2}.sin4x}\)\(=\dfrac{cos4x}{sin4x}\)

\(\Rightarrow\dfrac{cos4x}{sin4x}=\dfrac{1}{2}\)

\(\Leftrightarrow2cos4x=sin4x\)

\(\Leftrightarrow4.cos^24x=sin^24x\)

\(\Leftrightarrow4.cos^24x=1-cos^24x\)\(\Leftrightarrow cos^24x=\dfrac{1}{5}\Leftrightarrow\dfrac{1+cos8x}{2}=\dfrac{1}{5}\)

\(\Leftrightarrow cos8x=-\dfrac{3}{5}\)

Vậy..

Lời giải:

PT $\Leftrightarrow 2\sin 2x\cos 2x+2\cos 2x+4(\sin x+\cos x)=1+\cos ^22x-\sin ^22x=2\cos ^22x$

$\Leftrightarrow \sin 2x\cos 2x+\cos 2x+2(\sin x+\cos x)=\cos ^22x$

$\Leftrightarrow \cos 2x(\sin 2x+1-\cos 2x)+2(\sin x+\cos x)=0$

$\Leftrightarrow \cos 2x(2\sin x\cos x+2\sin ^2x)+2(\sin x+\cos x)=0$

$\Leftrightarrow \cos 2x\sin x(\cos x+\sin x)+(\sin x+\cos x)=0$

$\Leftrightarrow (\sin x+\cos x)(\cos 2x\sin x+1)=0$

Nếu $\sin x+\cos x=0$. Kết hợp $\sin ^2x+\cos ^2x=1$ suy ra $(\sin x, \cos x)=(\frac{1}{\sqrt{2}}; \frac{-1}{\sqrt{2}})$ và hoán vị

$\Rightarrow x=k\pi -\frac{\pi}{4}$ với $k$ nguyên.

Nếu $\cos 2x\sin x+1=0$

$\Leftrightarrow (1-2\sin ^2x)\sin x+1=0$

$\Leftrightarrow (1-\sin x)(2\sin ^2x+2\sin x+1)=0$

$\Rightarrow \sin x=1$

$\Rightarrow x=2k\pi +\frac{\pi}{2}$ với $k$ nguyên.

\(\text{1) }3sinx-4cosx=1\\ \Leftrightarrow cos^2x+\left(\frac{4cosx+1}{3}\right)^2=1\\ \Leftrightarrow cosx=\frac{-4\pm6\sqrt{6}}{25}\\ \\ \Leftrightarrow x=arccos\left(\frac{-4\pm6\sqrt{6}}{25}\right)+k2\pi\)

\(2\text{) }\sqrt{3}sinx-cosx=1\\ \Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sinx-sin\frac{\pi}{6}\cdot cosx=\frac{1}{2}\\ \Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin\frac{\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=\frac{\pi}{6}+a2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+a2\pi\\x=\pi+b2\pi\end{matrix}\right.\)

\(3\text{) }\sqrt{3}cosx+sinx=-2\\ \Leftrightarrow\frac{\sqrt{3}}{2}cosx+\frac{1}{2}sinx=-1\\ \Leftrightarrow sin\frac{\pi}{3}\cdot cosx+cos\frac{\pi}{3}\cdot sinx=-1\\ \Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=-1=sin\frac{3\pi}{2}\\ \\ \Leftrightarrow x+\frac{\pi}{3}=\frac{3\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{7\pi}{6}+k2\pi\)

\(4\text{) }cos4x-sin4x=1\\ \Leftrightarrow cos^24x+\left(cos4x-1\right)^2=1\\ \\ \Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+a\pi\\4x=b2\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{a\pi}{4}\\x=\frac{b\pi}{2}\end{matrix}\right.\)

\(5\text{) }\sqrt{3}cos4x+sin4x-2cos3x=0\\ \Leftrightarrow\frac{\sqrt{3}}{2}cos4x+\frac{1}{2}sin4x=cos3x\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos4x+sin\frac{\pi}{3}\cdot sin4x=cos3x\\ \Leftrightarrow cos\left(4x-\frac{\pi}{3}\right)=cos3x\\ \Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{3}=3x+a2\pi\\4x-\frac{\pi}{3}=-3x+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+a2\pi\\x=\frac{\pi}{21}+\frac{b2\pi}{7}\end{matrix}\right.\\ \Leftrightarrow x=\frac{\pi}{21}+\frac{k2\pi}{7}\)

\(6\text{) }cos^2x=3sin2x+3\\ \Leftrightarrow\frac{cos2x+1}{2}=3sin2x+3\)

Giải tương tự vd 1 và 4

7) Giải tương tự vd 1 và 4

\(P=\dfrac{-2sin5x.sinx-sinx}{2sin5x.cosx+cosx}=\dfrac{-sinx\left(2sin5x+1\right)}{cosx\left(2sin5x+1\right)}=-tanx\)

a.

\(\Leftrightarrow\dfrac{\sqrt{2}}{2}sin4x+\dfrac{\sqrt{2}}{2}cos4x=\dfrac{\sqrt{6}}{2}\)

\(\Leftrightarrow cos4x.cos\left(\dfrac{\pi}{4}\right)+sin4x.sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\)

\(\Leftrightarrow cos\left(4x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{6}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{4}=arccos\left(\dfrac{\sqrt{6}}{2}\right)+k2\pi\\4x-\dfrac{\pi}{4}=-arccos\left(\dfrac{\sqrt{6}}{2}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{1}{4}arccos\left(\dfrac{\sqrt{6}}{2}\right)+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{16}-\dfrac{1}{4}arccos\left(\dfrac{\sqrt{6}}{2}\right)+\dfrac{k\pi}{4}\end{matrix}\right.\)

b.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=\dfrac{\sqrt{3}}{6}\)

\(\Leftrightarrow cosx.cos\left(\dfrac{\pi}{3}\right)+sinx.sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{6}\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{6}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+arccos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\\x=\dfrac{\pi}{3}-arrcos\left(\dfrac{\sqrt{3}}{6}\right)+k2\pi\end{matrix}\right.\)

c.

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

\(\Leftrightarrow cosx.cos\left(\dfrac{\pi}{6}\right)+sinx.sin\left(\dfrac{\pi}{6}\right)=cos\left(\dfrac{3\pi}{4}\right)\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{6}\right)=cos\left(\dfrac{3\pi}{4}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{3\pi}{4}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11\pi}{12}+k2\pi\\x=-\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)

\(sin^8x-cos^8x-4sin^6x+6sin^4x-4sin^2x\)

\(=sin^8x-\left(1-sin^2x\right)^4-4sin^6x+6sin^4x-4sin^2x\)

\(=sin^8x-\left(1-4sin^2x+6sin^4x-4sin^6x+sin^8x\right)-4sin^6x+6sin^4x-4sin^2x\)\(=-1\) (bạn chép nhầm đề)

b/ \(\frac{sin6x+sin2x+sin4x}{1+cos2x+cos4x}=\frac{2sin4x.cos2x+sin4x}{1+cos2x+2cos^22x-1}=\frac{sin4x\left(2cos2x+1\right)}{cos2x\left(2cos2x+1\right)}=\frac{sin4x}{cos2x}=\frac{2sin2x.cos2x}{cos2x}=2sin2x\)

c/ \(\frac{1+sin2x}{cosx+sinx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{sin^2x+cos^2x+2sinx.cosx}{cosx+sinx}-\left(1-tan^2\frac{x}{2}\right)cos^2\frac{x}{2}\)

\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)=sinx+cosx-cosx=sinx\)

d/ \(cos4x+4cos2x+3=2cos^22x-1+4cos2x+3\)

\(=2\left(cos^22x+2cos2x+1\right)=2\left(cos2x+1\right)^2=2\left(2cos^2x-1+1\right)^2=8cos^4x\)

e/

\(A=\frac{cosx-cos3x+cos4x-cos2x}{sinx-sin3x+sin4x-sin2x}=\frac{2sin2x.sinx-2sin3x.sinx}{-2cos2x.sinx+2cos3x.sinx}\)

\(=\frac{sin2x-sin3x}{cos3x-cos2x}=\frac{-2cos\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}{-2sin\left(\frac{5x}{2}\right)sin\left(\frac{x}{2}\right)}=cot\left(\frac{5x}{2}\right)\)

\(B=sinx+2cos2x.sinx+2cos4x.sinx+2cos6x.sinx\)

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

\(=sin7x\)

a/

\(\sqrt{3}sin4x-cos4x=2cosx\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x-\frac{1}{2}cos4x=cosx\)

\(\Leftrightarrow sin\left(4x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{6}=\frac{\pi}{2}-x+k2\pi\\4x-\frac{\pi}{6}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{15}+\frac{k2\pi}{5}\\x=\frac{2\pi}{9}+\frac{k2\pi}{3}\end{matrix}\right.\)

b/

\(\Leftrightarrow cosx-\sqrt{3}sinx=sin2x-\sqrt{3}cos2x\)

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

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

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{6}-x\right)\)

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

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

c/

\(\Leftrightarrow cos3x-\sqrt{3}sin3x=\sqrt{3}cos2x-sin2x\)

\(\Leftrightarrow\frac{1}{2}cos3x-\frac{\sqrt{3}}{2}sin3x=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)

\(\Leftrightarrow cos\left(3x+\frac{\pi}{3}\right)=cos\left(2x+\frac{\pi}{6}\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)