Question

[Big City Boy]

Giải: (sin2x-cos2x)sinx+sin3x=(sinx+cosx)cosx

Answer 1

[Nguyễn Lê Phước Thịnh]

\(\Leftrightarrow sin2x\cdot sinx-cos2x\cdot sinx+sin2x\cdot cosx+sinx\cdot cos2x=cosx\left(sinx+cosx\right)\)

=>\(sin2x\left(sinx+cosx\right)=cosx\left(sinx+cosx\right)\)

=>\(\left(sinx+cosx\right)\cdot\left(sin2x-cosx\right)=0\)

=>\(cosx\cdot\left(2sinx-1\right)\cdot\sqrt{2}\cdot sin\left(x+\dfrac{pi}{4}\right)=0\)

=>\(\left[{}\begin{matrix}cosx=0\\2sinx-1=0\\sin\left(x+\dfrac{pi}{4}\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{pi}{2}+kpi\\sinx=\dfrac{1}{2}\\x+\dfrac{pi}{4}=kpi\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{pi}{2}+kpi\\x=-\dfrac{pi}{4}+kpi\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{pi}{2}+kpi\\x=-\dfrac{pi}{4}+kpi\\x=\dfrac{pi}{6}+k2pi\\x=\dfrac{5}{6}pi+k2pi\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{pi}{2}+kpi;-\dfrac{pi}{4}+kpi;\dfrac{pi}{6}+k2pi;\dfrac{5}{6}pi+k2pi\right\}\)