1.

\(3sin^22x-2sin2x.cos2x-4cos^22x=2\)

\(\Leftrightarrow-\dfrac{3}{2}\left(1-2sin^22x\right)-2sin2x.cos2x-2\left(2cos^22x-1\right)=\dfrac{5}{2}\)

\(\Leftrightarrow sin4x+\dfrac{7}{2}cos4x=-\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{53}}{2}\left(\dfrac{2}{\sqrt{53}}sin4x+\dfrac{7}{\sqrt{53}}cos4x\right)=-\dfrac{5}{2}\)

\(\Leftrightarrow sin\left(4x+arccos\dfrac{2}{\sqrt{53}}\right)=-\dfrac{5}{\sqrt{53}}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+arccos\dfrac{2}{\sqrt{53}}=arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\\4x+arccos\dfrac{2}{\sqrt{53}}=\pi-arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}+\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}-\dfrac{1}{4}arccos\dfrac{2}{\sqrt{53}}-\dfrac{1}{4}arcsin\left(-\dfrac{5}{\sqrt{53}}\right)+\dfrac{k\pi}{2}\end{matrix}\right.\)

2.

\(2\sqrt{3}cos^2x+6sinx.cosx=3+\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}\left(2cos^2x-1\right)+6sinx.cosx=3\)

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

\(\Leftrightarrow2\sqrt{3}\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)=3\)

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)

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

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

3.

\(3cos^24x+5sin^24x=2-2\sqrt{3}sin4x.cos4x\)

\(\Leftrightarrow4cos^24x+4sin^24x-cos^24x+sin^24x=2-2\sqrt{3}sin4x.cos4x\)

\(\Leftrightarrow4-cos8x=2-\sqrt{3}sin8x\)

\(\Leftrightarrow cos8x-\sqrt{3}sin8x=2\)

\(\Leftrightarrow\dfrac{1}{2}cos8x-\dfrac{\sqrt{3}}{2}sin8x=1\)

\(\Leftrightarrow cos\left(8x+\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow8x+\dfrac{\pi}{3}=k2\pi\)

\(\Leftrightarrow x=-\dfrac{\pi}{24}+\dfrac{k\pi}{4}\)

Bạn xem lại đề bài

a, \(cos2x+4cosx+1=0\)

\(\Leftrightarrow2cos^2x+4cosx=0\)

\(\Leftrightarrow2cosx\left(cosx+2\right)=0\)

\(\Leftrightarrow cosx=0\)

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

b, \(cos^22x=\dfrac{1}{4}\)

\(\Leftrightarrow4cos^22x-1=0\)

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

\(\Leftrightarrow cosx=\pm\dfrac{1}{2}\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{3}+k\pi\)

1.

\(2cos4x-3=0\)

\(\Leftrightarrow cos4x=\dfrac{3}{2}\)

Mà \(cos4x\in\left[-1;1\right]\)

\(\Rightarrow\) phương trình vô nghiệm.

2.

\(cos5x+2=0\)

\(\Leftrightarrow cos5x=-2\)

Mà \(cos5x\in\left[-1;1\right]\)

\(\Rightarrow\) phương trình vô nghiệm.

3.

\(cos2x+0,7=0\)

\(\Leftrightarrow cos2x=-\dfrac{7}{10}\)

\(\Leftrightarrow2x=\pm arccos\left(-\dfrac{7}{10}\right)+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{arccos\left(-\dfrac{7}{10}\right)}{2}+k\pi\)

4.

\(cos^22x-\dfrac{1}{4}=0\)

\(\Leftrightarrow cos^22x=\dfrac{1}{4}\)

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

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

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

A = 4 ( 2 sinx . cosx )² . cos²2x + cos²4x

A = 4 . sin²2x . cos²2x + cos²4x

A = ( 2 sin2x . cos2x)² + cos²4x

A = sin² 4x + cos²4x  = 1

1.

\(sin^3x+cos^3x=1-\dfrac{1}{2}sin2x\)

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

\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)=1-sinx.cosx\)

\(\Leftrightarrow\left(1-sinx.cosx\right)\left(sinx+cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx.cosx=1\\sinx+cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=2\left(vn\right)\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)

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

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

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

2.

\(\left|cosx-sinx\right|+2sin2x=1\)

\(\Leftrightarrow\left|cosx-sinx\right|-1+2sin2x=0\)

\(\Leftrightarrow\left|cosx-sinx\right|-\left(cosx-sinx\right)^2=0\)

\(\Leftrightarrow\left|cosx-sinx\right|\left(1-\left|cosx-sinx\right|\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\\left|cosx-sinx\right|=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\1-sin2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\sin2x=0\end{matrix}\right.\)

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

3.

\(2sin2x-3\sqrt{6}\left|sinx+cosx\right|+8=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^2-3\sqrt{6}\left|sinx+cosx\right|+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|sinx+cosx\right|=\sqrt{6}\left(vn\right)\\\left|sinx+cosx\right|=\dfrac{\sqrt{6}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left|sin\left(x+\dfrac{\pi}{4}\right)\right|=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\pm\dfrac{\sqrt{3}}{2}\)

...

giúp tớ với 

Đây cậu

1a.

Đặt \(5x+6=u\)

\(cos2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

1b.

Đặt \(2x+1=u\)

\(cos2u+3sinu=2\)

\(\Leftrightarrow1-2sin^2u+3sinu=2\)

\(\Leftrightarrow2sin^2u-3sinu+1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)

2a.

\(cos^2x-sin^2x+sin^2x+2cosx+1=0\)

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

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

\(\Leftrightarrow cosx=-1\)

\(\Leftrightarrow x=\pi+k2\pi\)