\(sin2x+cos2x=1\Rightarrow\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=1\)
\(\Rightarrow sin\left(2x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}=sin\dfrac{\pi}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\2x+\dfrac{\pi}{4}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
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