\(\dfrac{sin3x+cos3x}{1+2sin2x}=\dfrac{3sinx-4sin^3x+4cos^3x-3cosx}{1+2sin2x}\)
\(=\dfrac{\left(cosx-sinx\right)\left(4+2sin2x\right)-3\left(cosx-sinx\right)}{1+2sin2x}=\dfrac{\left(cosx-sinx\right)\left(1+2sin2x\right)}{1+2sin2x}=cosx-sinx\)
Pt trở thành:
\(sinx+cosx-sinx=\dfrac{3+cos2x}{5}\)
\(\Leftrightarrow5cosx=3+2cos^2x-1\)
\(\Leftrightarrow2cos^2x-5cosx+2=0\Rightarrow\left[{}\begin{matrix}cosx=2\left(loại\right)\\cosx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)
\(\Rightarrow x=\dfrac{\pi}{3};\dfrac{5\pi}{3}\)
Đúng 0
Bình luận (0)
