a)

\(4\sin (3x+\frac{\pi}{3})-2=0\Leftrightarrow \sin (3x+\frac{\pi}{3})=\frac{1}{2}=\sin (\frac{\pi}{6})\)

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

\(\Leftrightarrow \left[\begin{matrix} x=\frac{-\pi}{18}+\frac{2\pi}{3}\\ x=\frac{\pi}{6}+\frac{2\pi}{3}\end{matrix}\right.\) (k nguyên)

c)

\(\sin (x+\frac{x}{4})-1=0\Leftrightarrow \sin (\frac{5}{4}x)=1=\sin (\frac{\pi}{2})\)

\(\Rightarrow \frac{5}{4}x=\frac{\pi}{2}+2k\pi\Rightarrow x=\frac{2}{5}\pi+\frac{8}{5}k\pi \) (k nguyên)

d)

\(2\sin (2x+70^0)+1=0\Leftrightarrow \sin (2x+\frac{7}{18}\pi)=-\frac{1}{2}=\sin (\frac{-\pi}{6})\)

\(\Rightarrow \left[\begin{matrix} 2x+\frac{7}{18}\pi=\frac{-\pi}{6}+2k\pi\\ 2x+\frac{7}{18}\pi=\frac{7}{6}\pi+2k\pi\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x=\frac{-5\pi}{18}+k\pi\\ x=\frac{7}{18}\pi+k\pi\end{matrix}\right.\)

f)

\(\cos 2x-\cos 4x=0\)

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

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

b,e,g bạn xem lại đề, đơn vị không thống nhất.

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.

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

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

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

\(\Leftrightarrow cosx=-1\)

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

a. ĐKXĐ: ...

\(\frac{sinx}{cosx}+\frac{sin2x}{cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow\frac{sin2x.cosx+cos2x.sinx}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow sin3x\left(\frac{cosx.cos2x+cos3x}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{cosx\left(2cos^2x-1\right)+4cos^3x-3cosx}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{6cos^2x-4}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{3cos2x-1}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=\frac{1}{3}\end{matrix}\right.\)

b.

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

\(\Leftrightarrow4cos^32x-2cos2x-1=0\)

Pt bậc 3 này ko giải được, chắc bạn ghi nhầm đề

c. ĐKXĐ: ...

\(\frac{cosx}{sinx}-\frac{sinx}{cosx}=cosx-sinx\)

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

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

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

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

\(\Rightarrow t=\frac{t^2-1}{2}\Rightarrow t^2-2t-1=0\Rightarrow\left[{}\begin{matrix}t=1+\sqrt{2}\left(l\right)\\t=1-\sqrt{2}\end{matrix}\right.\)

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

d.

\(\Leftrightarrow2\left(cosx-cos3x\right)sin4x=sin3x\)

\(\Leftrightarrow2sin4x.cosx-2sin4x.cos3x=sin3x\)

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

\(\Leftrightarrow sin5x-sin7x-sinx=0\)

\(\Leftrightarrow-2cos6x.sinx-sinx=0\)

\(\Leftrightarrow sinx\left(2cos6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos6x=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

a/

\(cos^6x+sin^2x=1\)

\(\Leftrightarrow cos^6x-\left(1-sin^2x\right)=0\)

\(\Leftrightarrow cos^6x-cos^2x=0\)

\(\Leftrightarrow cos^2x\left(cos^4x-1\right)=0\)

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

\(\Leftrightarrow-cos^2x.sin^2x=0\)

\(\Leftrightarrow sin^22x=0\)

\(\Leftrightarrow sin2x=0\)

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

b/

\(cos^6x-sin^6x=\frac{13}{18}cos^22x\)

\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(cos^4x+sin^4x+sin^2x.cos^2x\right)=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left[\left(sin^2x+cos^2x\right)^2-sin^2x.cos^2x\right]=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left(1-\frac{1}{4}sin^22x\right)=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left(1-\frac{1}{4}\left(1-cos^22x\right)\right)=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left(\frac{3}{4}+\frac{1}{4}cos^22x\right)=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left(\frac{1}{4}cos^22x-\frac{13}{18}cos2x+\frac{3}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\\frac{1}{4}cos^22x-\frac{13}{18}cos2x+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

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

c/

\(cos^4x+sin^6x=cos2x\)

\(\Leftrightarrow\left(\frac{1+cos2x}{2}\right)^2+\left(\frac{1-cos2x}{2}\right)^3=cos2x\)

\(\Leftrightarrow cos^32x-5cos^2x+7cos2x-3=0\)

\(\Leftrightarrow\left(cos2x-1\right)^2\left(cos2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=3\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=k2\pi\)

\(\Rightarrow x=k\pi\)

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.

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

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

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

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

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

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

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

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

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

\(0< =sin^2x< =1\)

=>\(-2< =sin^2x-2< =-1\)

=>\(sin^2x-2< 0\)

\(0< =cos^2x< =1\)

=>\(-2< =cos^2x-2< =-1\)

\(\Leftrightarrow cos^2x-2< 0\)

\(\sqrt{sin^4x+4cos^2x}+\sqrt{cos^4x+4\cdot sin^2x}\)

\(=\sqrt{sin^4x+4\left(1-sin^2x\right)}+\sqrt{cos^4x+4\cdot\left(1-cos^2x\right)}\)

\(=\sqrt{sin^4x-4sin^xx+4}+\sqrt{cos^4x-4\cdot cos^2x+4}\)

\(=\sqrt{\left(sin^2x-2\right)^2}+\sqrt{\left(cos^2x-2\right)^2}\)

\(=\left|sin^2x-2\right|+\left|cos^2x-2\right|\)

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

\(=4-\left(sin^2x+cos^2x\right)=4-1=3\)