a/

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x-2\left(1-sin^22x\right)=0\)

\(\Leftrightarrow1-\frac{1}{2}\left(cos6x+cos2x\right)-2cos^22x=0\)

\(\Leftrightarrow1-cos4x.cos2x-2cos^22x=0\)

\(\Leftrightarrow2cos^22x-1+cos4x.cos2x=0\)

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

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

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

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

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

b/

\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}\right)+\sqrt{3}sin2x=1\)

\(\Leftrightarrow cos2x.cos\left(\frac{\pi}{3}\right)-sin2x.sin\left(\frac{\pi}{3}\right)+\frac{1}{2}+\sqrt{3}sin2x=1\)

\(\Leftrightarrow\frac{1}{2}cos2x+\frac{\sqrt{3}}{2}sin2x=\frac{1}{2}\)

\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=\frac{1}{2}\)

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

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

c/

\(\Leftrightarrow5+5cosx=2+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)

\(\Leftrightarrow3+5cosx=sin^2x-cos^2x\)

\(\Leftrightarrow3+5cosx=1-cos^2x-cos^2x\)

\(\Leftrightarrow2cos^2x+5cosx+2=0\)

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

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

5.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)

\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)

\(\Leftrightarrow sin^22x=\frac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)

6.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)

\(\Leftrightarrow-3sin^22x+sin2x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)

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

1.

\(\Rightarrow4cos^3x.cos3x+4sin^3x.sin3x=\sqrt{2}\)

\(\Leftrightarrow\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x=\sqrt{2}\)

\(\Leftrightarrow3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+cos6x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+4cos^32x-3cos2x=\sqrt{2}\)

\(\Leftrightarrow4cos^32x=\sqrt{2}\)

\(\Leftrightarrow cos2x=\frac{\sqrt{2}}{2}\)

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

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

a/

\(\Leftrightarrow4sin^3x+6\sqrt{2}sinx.cosx-8sinx=0\)

\(\Leftrightarrow2sinx\left(2sin^2x+3\sqrt{2}cosx-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\2sin^2x+3\sqrt{2}cosx-4=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\left(1-cos^2x\right)+3\sqrt{2}cosx-4=0\)

\(\Leftrightarrow-2cos^2x+3\sqrt{2}cosx-2=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=\sqrt{2}>1\left(l\right)\\cosx=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow x=\pm\frac{\pi}{4}+k2\pi\)

b/

\(\Leftrightarrow4cos^3x+8sinx.cosx-7cosx=0\)

\(\Leftrightarrow cosx\left(4cos^2x+8sinx-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\4cos^2x+8sinx-7=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow4\left(1-sin^2x\right)+8sinx-7=0\)

\(\Leftrightarrow-4sin^2x+8sinx-3=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=\frac{3}{2}\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)

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

c/

ĐKXĐ; ...

\(\Leftrightarrow\frac{sinx}{cosx}+\frac{cosx}{sinx}-5+\frac{3}{sin^22x}=0\)

\(\Leftrightarrow\frac{sin^2x+cos^2x}{sinx.cosx}-5+\frac{3}{sin^22x}=0\)

\(\Leftrightarrow\frac{3}{sin^22x}+\frac{2}{sin2x}-5=0\)

Đặt \(\frac{1}{sin2x}=t\Rightarrow3t^2+2t-5=0\)

\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{5}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{1}{sin2x}=1\\\frac{1}{sin2x}=-\frac{5}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{3}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(-\frac{3}{5}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(-\frac{3}{5}\right)+k\pi\end{matrix}\right.\)

\(A=2(\sin ^6x+\cos ^6x)-3(\sin ^4x+\cos ^4x)\)

\(=2(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-3(\sin ^4x+\cos ^4x)\)

\(=2(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-3(\sin ^4x+\cos ^4x)\)

\(=-(\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x)=-(\sin ^2x+\cos ^2x)^2=-1^2=-1\)

là giá trị không phụ thuộc vào biến (đpcm)

-----------------------

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

\(=(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-2\sin ^4x-\cos ^4x+\sin ^2x\)

\(=\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x-2\sin ^4x-\cos ^4x+\sin ^2x\)

\(=-\sin ^4x-\sin ^2x\cos ^2x+\sin ^2x=-\sin ^2x(\sin ^2x+\cos ^2x)+\sin ^2x\)

\(=-\sin ^2x+\sin ^2x=0\)

là giá trị không phụ thuộc vào biến (đpcm)

\(C=(\sin ^4x+\cos ^4x-1)(\tan ^2x+\cot ^2x+2)=(\sin ^4x+\cos ^4x-1)(\frac{\sin ^2x}{\cos ^2x}+\frac{\cos ^2x}{\sin ^2x}+2)\)

\(=(\sin ^4x+\cos ^4x-1).\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin ^2x\cos ^2x}=(\sin ^4x+\cos ^4x-1).\frac{(\sin ^2x+\cos ^2x)^2}{\sin ^2x\cos ^2x}\)

\(=(\sin ^4x+\cos ^4x-1).\frac{1}{\sin ^2x\cos ^2x}=\frac{(\sin ^2x)^2+(\cos ^2x)^2+2\sin ^2x\cos ^2x-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}\)

\(=\frac{(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}=\frac{1-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}=\frac{-2\sin ^2x\cos ^2x}{\sin ^2x\cos ^2x}=-2\)

là giá trị không phụ thuộc vào biến $x$

--------------------

\(D=\frac{1}{\cos ^6x}-\tan ^6x-\frac{\tan ^2x}{\cos ^2x}=\frac{1}{\cos ^6x}-\frac{\sin ^6x}{\cos ^6x}-\frac{\sin ^2x}{\cos ^4x}\)

\(=\frac{1-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{(\sin ^2x+\cos ^2x)^3-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)

\(=\frac{\sin ^6x+\cos ^6x+3\sin ^2x\cos ^2x(\sin ^2x+\cos ^2x)-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)

\(=\frac{\cos ^6x+3\sin ^2x\cos ^2x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{\cos ^4x+2\sin ^2x}{\cos ^4x}\)

\(=1+\frac{2\sin ^2x}{\cos ^4x}\)

Giá trị biểu thức này vẫn phụ thuộc vào $x$. Bạn xem lại đề.

\(\frac{sin3x+sinx+sin4x}{cos4x+1+cosx+cos3x}=\frac{2sin2x.cosx+2sin2x.cos2x}{2cos^22x+2cos2x.cosx}=\frac{2sin2x\left(cosx+cos2x\right)}{2cos2x\left(cos2x+cosx\right)}=\frac{sin2x}{cos2x}=tan2x\)

\(\frac{sin^22x+2cos\left(2\pi+\pi+2x\right)-2}{-3+4cos2x+cos\left(\pi-4x\right)}=\frac{sin^22x-2cos2x-2}{-3+4cos2x-cos4x}=\frac{4sin^2x.cos^2x-2\left(2cos^2x-1\right)-2}{-3+4\left(1-2sin^2x\right)-\left(1-2sin^22x\right)}\)

\(=\frac{4cos^2x\left(sin^2x-1\right)}{-8sin^2x+2sin^22x}=\frac{2cos^2x.\left(-cos^2x\right)}{-4sin^2x+4sin^2x.cos^2x}=\frac{cos^4x}{2sin^2x\left(1-cos^2x\right)}\)

\(=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)

\(A=\left(sin^2x+cos^2x\right)\left(sin^4x+cos^4x-sin^2x.cos^2x\right)+m\left(sin^4x+cos^4x\right)+\left(m+1\right)sin^22x\)

\(=\left(sin^2x+cos^2x\right)^2-3sin^2x.cos^2x+m\left[\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right]+\left(m+1\right)sin^22x\)

\(=1-\frac{3}{4}sin^22x+m\left(1-\frac{1}{2}sin^22x\right)+\left(m+1\right)sin^22x\)

\(=m+1+\left(-\frac{3}{4}-\frac{m}{2}+m+1\right)sin^22x\)

\(=m+1+\left(\frac{m}{2}+\frac{1}{4}\right)sin^22x\)

Để biểu thức ko phụ thuộc x \(\Rightarrow\frac{m}{2}+\frac{1}{4}=0\Rightarrow m=-\frac{1}{2}\)