ĐKXĐ: \(sin2x\ne-\dfrac{1}{2}\)
Ta có: \(sin3x+cos3x=3sinx-4sin^3x+4cos^3x-3cosx\)
\(=\left(cosx-sinx\right)\left(4+2sin2x\right)-3\left(cosx-sinx\right)\)
\(=\left(cosx-sinx\right)\left(1+2sin2x\right)\)
Nên pt trở thành:
\(5\left(sinx+\dfrac{\left(cosx-sinx\right)\left(1+2sin2x\right)}{1+2sin2x}\right)=cos2x+3\)
\(\Leftrightarrow5cosx=cos2x+3\)
\(\Leftrightarrow2cos^2x-5cosx+2=0\Rightarrow\left[{}\begin{matrix}cosx=2\left(loại\right)\\cosx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)
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