\(d=\dfrac{\left(cos^4\left(x\right)-tanx\right)}{cos\left(2x\right)}:t=\dfrac{dcos\left(2x\right)-cos^4\left(x\right)}{anx};cos\left(2x\right)\ne0,n\ne0\)

\(\left\{{}\begin{matrix}d=\dfrac{\left(cos^4\left(x\right)-tanx\right)}{cos\left(2x\right)}\\\dfrac{cos^4\left(x\right)-tanx}{cos\left(2x\right)}=d\end{matrix}\right.\)  

\(\Leftrightarrow\) \(\dfrac{\left(cos^4\left(x\right)-tanx\right)cos\left(2x\right)}{cos\left(2x\right)}=dcos\left(2x\right);cos\left(2x\right)\ne0\)

\(\Leftrightarrow\) \(cos^4\left(x\right)-antx=dcos\left(2x\right);cos\left(2x\right)\ne0\)

\(\Leftrightarrow\) \(cos^4\left(x\right)-antx-cos^4\left(x\right)=dcos\left(2x\right)-cos^4\left(x\right);cos\left(2x\right)\ne0\)

\(\Leftrightarrow\) \(-antx=dcos\left(2x\right)-cos^4\left(x\right);cos\left(2x\right)\ne0\)

\(\Leftrightarrow\) \(\dfrac{-antx}{-anx}=\dfrac{dcos\left(2x\right)}{-anx}-\dfrac{cos^4\left(x\right)}{-anx};cos\left(2x\right)\ne0,n\ne0\)

\(\Rightarrow\) \(suy\) \(ra\) \(:\) \(d=-\dfrac{dcos\left(2x\right)-cos^4\left(x\right)}{-anx};cos\left(2x\right)\ne0,n\ne0\)

a.

\(\Leftrightarrow\pi cos2x=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow cos2x=\frac{1}{2}+2k\)

Do \(-1\le cos2x\le1\Rightarrow-1\le\frac{1}{2}+2k\le1\)

\(\Rightarrow k=0\)

\(\Rightarrow cos2x=\frac{1}{2}\)

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

b.

\(\Leftrightarrow cos4x=1\)

\(\Leftrightarrow4x=k2\pi\)

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

c.

ĐKXĐ: \(cosx\ne1\)

\(\Leftrightarrow cos2x-1=1-cosx\)

\(\Leftrightarrow2cos^2x-1-1=1-cosx\)

\(\Leftrightarrow2cos^2x+cosx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\left(l\right)\\cosx=-\frac{3}{2}< -1\left(l\right)\end{matrix}\right.\)

Vậy pt đã cho vô nghiệm

d.

ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\tanx\ne1\end{matrix}\right.\)

\(\Leftrightarrow cos2x=tanx-1\)

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

\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx\right)=\frac{cosx-sinx}{-cosx}\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\Leftrightarrow tanx=1\left(l\right)\\cosx+sinx=-\frac{1}{cosx}\left(1\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos2x+\frac{1}{2}sin2x=-1\)

\(\Leftrightarrow cos2x+sin2x=-3\)

Do \(\left\{{}\begin{matrix}cos2x\ge-1\\sin2x\ge-1\end{matrix}\right.\) \(\Rightarrow cos2x+sin2x\ge-2>-3\)

\(\Rightarrow\left(1\right)\) vô nghiệm

Vậy pt đã cho vô nghiệm

\(\frac{sin4x-sin2x}{1-cos2x+cos4x}=\frac{2sin2x.cos2x-sin2x}{1-cos2x+2cos^22x-1}=\frac{sin2x\left(2cos2x-1\right)}{cos2x\left(2cos2x-1\right)}=\frac{sin2x}{cos2x}=tan2x\)

\(\Rightarrow\) đề sai

b/

\(\frac{1-cos4x}{sin4x}=\frac{1-\left(1-2sin^22x\right)}{2sin2x.cos2x}=\frac{2sin^22x}{2sin2x.cos2x}=\frac{sin2x}{cos2x}=tan2x\)

Đề sai tiếp lần 2

a/ ĐKXĐ:...

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

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

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

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

\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=1\)

\(\Leftrightarrow x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\frac{3\pi}{4}+k2\pi\)

b/

ĐKXĐ: ...

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x-1\right)+cos4x\left(2sinx-1\right)=0\)

\(\Leftrightarrow2sinx.sin4x-2sinx-sin4x+1+2sinx.cos4x-cos4x=0\)

\(\Leftrightarrow2sinx\left(sin4x+cos4x\right)-\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin4x+cos4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\4x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\4x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\left(l\right)\end{matrix}\right.\)

c/

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

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

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

d/

\(\Leftrightarrow sin2x-2cos2x-5=2sin2x-cos2x-6\)

\(\Leftrightarrow sin2x+cos2x=1\)

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

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

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

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

c/

Hình như câu này đề sai

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

\(\Leftrightarrow sinx+cosx-\left(sinx+cosx\right)=\sqrt{2}\)

\(\Leftrightarrow0=\sqrt{2}\)

Pt vô nghiệm

d/ Hình như câu này đề cũng sai

\(\Leftrightarrow sin2x-2cos2x-5=0\)

\(\Leftrightarrow\frac{1}{\sqrt{5}}sin2x-\frac{2}{\sqrt{5}}cos2x=\sqrt{5}\)

\(\Leftrightarrow sin\left(2x-a\right)=\sqrt{5}\) (với \(sina=\frac{2}{\sqrt{5}};cosa=\frac{1}{\sqrt{5}}\))

Pt vô nghiệm do \(\sqrt{5}>1\)

a. cos2x + cos4x + cos6x = 0

\(\Leftrightarrow\left(cos2x+cos6x\right)+cos4x=0\\ \Leftrightarrow2cos4x.cos2x+cos4x=0\\ \Leftrightarrow cos4x\left(2cos2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\left(k\in Z\right)}\)

1.

\(cos2x+cos6x+cos4x=0\)

\(\Leftrightarrow2cos4x.cos2x+cos4x=0\)

\(\Leftrightarrow cos4x\left(2cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)

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

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

2.

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

\(\Leftrightarrow1+2cos^2x-1+2cos2x.cosx=0\)

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

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

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

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

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