Question

sin²⁰¹⁸x + cos²⁰¹⁸x = 1. Giải phương trình lượng giác

Answer 1

pt<=>sin²⁰¹⁸x+cos²⁰¹⁸x=sin²x+cos²x
<=>sin²x.(sin²⁰¹⁶x-1)=cos²x.(1-cos²⁰¹⁶x)

Ta có:\(\left\{{}\begin{matrix}sin^2x\ge0\\cos^2x\ge0\end{matrix}\right.\)và\(\left\{{}\begin{matrix}sin^{2016}x-1\ge0\\1-cos^{2016}x\le0\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}VT\ge0\\VP\le0\end{matrix}\right.\)

=>VT=VP=0

<=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}sin^2x=0\\sin^{2016}x=1\end{matrix}\right.\\\left[{}\begin{matrix}cos^2x=0\\cos^{2016}x=1\end{matrix}\right.\end{matrix}\right.\)<=>x=\(\dfrac{k\Pi}{2}\)