giải các phương trình : 1) cos3x + cos2x -cosx -1=0 2) (2cosx - 1). (2sinx + cosx) = sin2x - sinx 3) sinx + cosx = 0

1) cos3x + cos2x -cosx -1=0

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Leftrightarrow cos3x-cosx+cos2x-1=0\\
\Leftrightarrow -2sin2x.sinx-2sin^{2} x=0\\
\Leftrightarrow sinx( sin2x+sinx) =0\\
\Leftrightarrow sinx=0
\end{array}$

hoặc sin2x+sinx=0

TH1: sinx=0 ⟹ x=k$\displaystyle \pi $

TH2: sin2x+sinx=0 ⟹ x=± $\displaystyle \frac{2\pi }{3} +k2\pi $

3) $\displaystyle sinx+cosx=0$

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Leftrightarrow \sqrt{2} sin\left( x+\frac{\pi }{4}\right) =0\\
\Leftrightarrow sin\left( x+\frac{\pi }{4}\right) =0\\
\Leftrightarrow x+\frac{\pi }{4} =k\pi \\
\Leftrightarrow x=k\pi -\frac{\pi }{4}
\end{array}$

2)

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\ ( 2cosx-1)( 2sinx+cosx) =sin2x-sinx\\
\Leftrightarrow ( 2cosx-1)( 2sinx+cosx) =2sinxcosx-sinx\\
\Leftrightarrow ( 2cosx-1)( 2sinx+cosx-sinx) =0\\
\Leftrightarrow ( 2cosx-1)( sinx+cosx) =0
\end{array}$

TH1:

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
2cosx-1=0\\
\Leftrightarrow cosx=\frac{1}{2}\\
\Leftrightarrow x=\pm \frac{\pi }{3} +k2\pi 
\end{array}$

TH2:

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
sinx+cosx=0\\
\Leftrightarrow x=k\pi -\frac{\pi }{4}
\end{array}$