a: \(\Leftrightarrow2\cdot cos2x\left(sinx-2\right)+sinx-2=0\)
=>(sin x-2)(2*cos2x+1)=0
=>2*cos2x+1=0
=>cos2x=-1/2
=>2x=2/3pi+k2pi hoặc 2x=-2/3pi+k2pi
=>x=1/3pi+kpi hoặc x=-1/3pi+kpi
b:
ĐKXĐ: \(2\sqrt{2}\cdot sin\left(\dfrac{x}{2}\right)\cdot cos\left(\dfrac{x}{2}\right)+1< >0\)
=>\(\sqrt{2}\cdot sinx+1< >0\)
=>\(sinx< >-\dfrac{1}{2}\)
=>\(\left\{{}\begin{matrix}x< >-\dfrac{pi}{4}+k2pi\\x< >\dfrac{5}{4}pi+k2pi\end{matrix}\right.\)
PT\(\Rightarrow sin2x+cos2x-sinx-cosx+1=0\)
\(\Leftrightarrow\left(sin2x-sinx\right)+\left(cos2x-cosx+1\right)=0\)
=>\(sinx\left(2\cdot cosx-1\right)+\left(2cos^2x-1-cosx+1\right)=0\)
=>\(sinx\left(2\cdot cosx-1\right)+cosx\left(2cosx-1\right)=0\)
=>\(\left(2cosx-1\right)\left(sinx+cosx\right)=0\)
=>\(\sqrt{2}sin\left(x+\dfrac{pi}{4}\right)\cdot\left(2cosx-1\right)=0\)
=>\(\left[{}\begin{matrix}sin\left(x+\dfrac{pi}{4}\right)=0\\2cosx-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{pi}{4}=kpi\\cosx=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{pi}{4}+kpi\\x=\pm\dfrac{pi}{3}+k2pi\end{matrix}\right.\)
Kết hợp ĐKXĐ, ta được:
\(\left[{}\begin{matrix}x=\pm\dfrac{pi}{3}+k2pi\\x=\dfrac{3}{4}pi+k2pi\end{matrix}\right.\)
