\(2cosx+3sinx+\dfrac{1}{3}cos2x-sin2x=\dfrac{8}{3}\)
\(\Leftrightarrow6cosx+9sinx+cos2x-3sin2x-8=0\)
\(\Leftrightarrow6cosx-6sinx.cosx+9sinx-9+1+1-2sin^2x=0\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-9\left(1-sinx\right)+2\left(1-sinx\right)\left(1+sinx\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(6cosx-9+2+2sinx=0\right)\)
\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-sinx=0\\6cosx+2sinx=7\end{matrix}\right.\)
TH1: \(1-sinx=0\Rightarrow sinx=1\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)
TH2: \(6cosx+2sinx=7\) (1)
Do \(6^2+2^2=40< 7^2=49\Rightarrow\) pt (1) vô nghiệm
Vậy pt đã cho có nghiệm \(x=\dfrac{\pi}{2}+k2\pi\)
