Đáp án:
\(\left[ \begin{array}{l}x=k\pi\\x=\frac{-\pi}{4}+k\pi\end{array} \right.\) (k∈Z)
Giải thích các bước giải:
2.cos³x=√2.sin(x+$\frac{\pi}{4}$) + cosx
<-> 2.cos³x=sinx+cosx+cosx
<-> 2cos³x-2cosx-sinx=0
<-> 2cosx(cos²x-1)-sinx=0
<-> -2cosx.sin²x-sinx=0
<-> sinx(sin2x+1)=0
<-> \(\left[ \begin{array}{l}sinx=0\\sin2x=-1\end{array} \right.\)
<-> \(\left[ \begin{array}{l}x=k\pi\\2x=\frac{-\pi}{2}+k2\pi\end{array} \right.\)
<-> \(\left[ \begin{array}{l}x=k\pi\\x=\frac{-\pi}{4}+k\pi\end{array} \right.\) (k∈Z)
