\(5\left(sinx+\dfrac{cos3x+sin3x}{1+2sin2x}\right)=cos2x+3\) (1)
\(Đk:\) \(1+sin2x\ne0\Rightarrow sin2x\ne-\dfrac{1}{2}=sin\left(-\dfrac{\pi}{6}\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne-\dfrac{\pi}{12}+m2\pi\\x\ne\dfrac{7\pi}{12}+n2\pi\end{matrix}\right.\) \(\left(m,n\in Z\right)\)
Ta có: \(VT_{\left(1\right)}=5\cdot\dfrac{sinx+2sin2x\cdot sinx+cos3x+sin3x}{1+2sin2x}\)
\(=5\cdot\dfrac{sinx+(cosx-cos3x)+cos3x+sin3x}{1+2sin2x}\)
\(=5\cdot\dfrac{\left(sin3x+sinx\right)+cosx}{1+2sin2x}\)
\(=5\cdot\dfrac{2sin2x\cdot cosx+cosx}{1+2sinx}\)
\(=5\cdot\dfrac{\left(2sinx+1\right)\cdot cosx}{1+2sin2x}\)
\(=5cosx\)
Từ Pt (1) \(\Rightarrow5cosx=cos2x+3\)
\(\Rightarrow2cos^2x-1+3-5cosx=0\)
\(\Rightarrow2cos^2x-5cosx+2=0\)
\(\Rightarrow\left\{{}\begin{matrix}cosx=2\left(l\right)\\cosx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow cosx=\dfrac{1}{2}\) \(\Rightarrow x=\pm\dfrac{\pi}{3}+k2\pi\left(k\in Z\right)\)(tm đk của x)
Với \(x=\dfrac{\pi}{3}+k2\pi\)\(\in\left(0;2\pi\right)\)\(\Rightarrow0< \dfrac{\pi}{3}+k2\pi< 2\pi\Rightarrow-\dfrac{1}{6}< k< \dfrac{5}{6}\)
\(\Rightarrow k=0\) \(\Rightarrow x=\dfrac{\pi}{3}\)
Làm tương tự với trường hợp \(x=-\dfrac{\pi}{3}+k2\pi\in\left(0;2\pi\right)\)
ta được nghiệm \(x=\dfrac{5\pi}{3}\)
Vậy phương trình đã cho có hai nghiệm \(x=\dfrac{\pi}{3}\) và \(x=\dfrac{5\pi}{3}\)
thỏa mãn đk \(x\in\left(0;2\pi\right)\).