\(\pi< a< \frac{3\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa< 0\end{matrix}\right.\) \(\Rightarrow sin2a=2sina.cosa>0\)

\(\Rightarrow sin2a=\sqrt{1-cos^22a}=\frac{3\sqrt{7}}{8}\)

\(cos2a=1-2sin^2a=\frac{1}{8}\)

\(\Leftrightarrow sin^2a=\frac{7}{16}\Rightarrow sina=-\frac{\sqrt{7}}{4}\)

\(\Rightarrow M=\frac{-\frac{\sqrt{7}}{4}-\frac{3\sqrt{7}}{8}}{-\frac{\sqrt{7}}{4}+\frac{3\sqrt{7}}{8}}=...\)

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

\(=sinx\left(\frac{1+cosx-\left(1-cosx\right)}{1+cosx}\right)=\frac{2sinx.cosx}{1+cosx}\)

\(1-sin2x.sin3x-cos2x.cos3x=1-\left(cos3x.cos2x+sin3x.sin2x\right)=1-cos\left(3x-2x\right)=1-cosx\)

\(\Rightarrow\frac{1-sin2x.sin3x-cos2x.cos3x}{sinx\left(1-tan^2\frac{x}{2}\right)}=\frac{1-cosx}{\frac{2sinx.cosx}{1+cosx}}=\frac{\left(1-cosx\right)\left(1+cosx\right)}{2sinx.cosx}\)

\(=\frac{1-cos^2x}{2sinx.cosx}=\frac{sin^2x}{2sinx.cosx}=\frac{sinx}{2cosx}=\frac{1}{2}tanx\)

Xem lại đề bài đi

Đề sai nhiều chỗ vậy, lần sau ghi đúng đề đi.

\(cos3x+sin7x=2sin^2\left(\dfrac{\pi}{4}-\dfrac{5x}{2}\right)+2cos^2\dfrac{9x}{2}\)

\(\Leftrightarrow cos3x+sin7x=cos\left(\dfrac{\pi}{2}-5x\right)+1-2cos^2\dfrac{9x}{2}\)

\(\Leftrightarrow cos3x+sin7x=sin5x-cos9x\)

\(\Leftrightarrow2cos6x.cos3x+2cos6x.sinx=0\)

\(\Leftrightarrow2cos6x.\left(cos3x+sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\cos3x+sinx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\cos3x+cos\left(\dfrac{\pi}{2}-x\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\2cos\left(\dfrac{\pi}{4}+x\right).cos\left(2x-\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\cos\left(\dfrac{\pi}{4}+x\right)=0\\cos\left(2x-\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+\dfrac{k\pi}{6}\\x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{3\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)

a: ĐKXĐ: sin 2x<>1

=>2x<>pi/2+k2pi

=>x<>pi/4+kpi

\(\dfrac{cos2x}{sin2x-1}=0\)

=>cos2x=0

=>2x=pi/2+kpi

=>x=pi/4+kpi/2

Kết hợp ĐKXĐ, ta được:

x=3/4pi+k2pi hoặc x=7/4pi+k2pi

b: cos(sinx)=1

=>sin x=kpi

=>sin x=0

=>x=kpi

c: \(2\cdot sin^2x-1+cos3x=0\)

=>cos3x+cos2x=0

=>cos3x=-cos2x=-sin(pi/2-2x)=sin(2x-pi/2)

=>cos3x=cos(pi/2-2x+pi/2)=cos(pi-2x)

=>3x=pi-2x+k2pi hoặc 3x=-pi+2x+k2pi

=>x=-pi+k2pi hoặc x=pi/5+k2pi/5

e: cos3x=-cos7x

=>cos3x=cos(pi-7x)

=>3x=pi-7x+k2pi hoặc 3x=-pi+7x+k2pi

=>x=pi/10+kpi/5 hoặc x=pi/4-kpi/2

a: \(\Leftrightarrow\dfrac{1}{2}\cdot\cos2x\cdot\cos x-\cos2x=0\)

\(\Leftrightarrow\cos2x=0\)

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

hay \(x=\dfrac{\Pi}{4}+\dfrac{k\Pi}{2}\)

b: \(\Leftrightarrow\dfrac{1}{2}\cdot\left[\cos\left(5x-x\right)-\cos\left(5x+x\right)\right]=\dfrac{1}{2}\cdot\left[\cos\left(3x-2x\right)-\cos5x\right]\)

\(\Leftrightarrow\cos4x-\cos6x=\cos x-\cos5x\)

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

\(\left\{{}\begin{matrix}-1\le cos7x\le1\\-1\le sin2x\le1\end{matrix}\right.\)

\(\Rightarrow cos7x.sin2x\ge-1\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}cos7x=-1\\sin2x=1\end{matrix}\right.\\\left\{{}\begin{matrix}cos7x=1\\sin2x=-1\end{matrix}\right.\end{matrix}\right.\) (ko tồn tại x thỏa mãn)

Vậy pt vô nghiệm

\(\Leftrightarrow cosx-cos7x-3\sqrt{3}sinx=0\)

\(\Leftrightarrow2sin4x.sin3x-3\sqrt{3}sinx=0\)

\(\Leftrightarrow2sin4x.\left(3sinx-4sin^3x\right)-3\sqrt{3}sinx=0\)

\(\Leftrightarrow sinx\left(6sin4x-8sin^2x.sin4x-3\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=...\\6sin4x-8sin^2x.sin4x=3\sqrt{3}\left(1\right)\end{matrix}\right.\)

Xét \(\left(1\right)\Leftrightarrow6sin4x-4sin4x\left(1-cos2x\right)=3\sqrt{3}\)

\(\Leftrightarrow2sin4x+4sin4x.cos2x=3\sqrt{3}\)

\(\Leftrightarrow sin4x+4sin2x.cos^22x=\frac{3\sqrt{3}}{2}\)

Ta có:

\(1=sin^22x+\frac{cos^22x}{2}+\frac{cos^22x}{2}\ge3\sqrt[3]{\frac{\left(sin2x.cos^22x\right)^2}{4}}\)

\(\Rightarrow\left(sin2x.cos^22x\right)^2\le\frac{4}{27}\Rightarrow sin2x.cos^22x\le\frac{2\sqrt{3}}{9}\)

\(\Rightarrow sin4x+4sin2x.cos^22x\le1+\frac{8\sqrt{3}}{9}< \frac{3\sqrt{3}}{2}\) nên pt vô nghiệm

a/

\(\Leftrightarrow2sin4x.cos3x=2sin7x.cos3x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\sin7x=sin4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\frac{\pi}{2}+k\pi\\7x=4x+k2\pi\\7x=\pi-4x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k\pi}{3}\\x=\frac{k2\pi}{3}\\x=\frac{\pi}{11}+\frac{k2\pi}{11}\end{matrix}\right.\)

b.

\(\Leftrightarrow2cos4x.cosx=2cos8x.cosx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos8x=cos4x\end{matrix}\right.\)

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

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