Question

[Nguyễn Minh Ngọc]

Giải phương trình đối xứng: cosx + sinx = 2sinx.cosx.(sinx + cosx)

Answer 1

[Nguyễn Lê Phước Thịnh]

=>(cosx+sinx)-2*sinx*cosx*(sinx+cosx)=0

=>\(\left(sinx+cosx\right)\left(2\cdot sinx\cdot cosx-1\right)=0\)

=>\(\sqrt{2}\cdot sin\left(x+\dfrac{pi}{4}\right)\cdot\left(sin2x-1\right)=0\)

=>\(\left[{}\begin{matrix}sin\left(x+\dfrac{pi}{4}\right)=0\\sin2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{pi}{4}=kpi\\sin2x=1\end{matrix}\right.\)

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