2. Ta có : \(\sin x-\sqrt{3}\cos x=1\)
\(\Leftrightarrow\dfrac{1}{2}\sin x-\dfrac{\sqrt{3}}{2}\cos x=\dfrac{1}{2}\)
\(\Leftrightarrow\cos60.\sin x-\sin60.\cos x=\dfrac{1}{2}\)
\(\Leftrightarrow\sin\left(\dfrac{\pi}{3}-x\right)=-\dfrac{1}{2}\)
\(\Leftrightarrow-x+\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{2}-k2\pi\)
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4. Tương tự câu 2 ta được : \(\sin\left(\dfrac{\pi}{3}+4x\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow4x+\dfrac{\pi}{3}=\dfrac{\pi}{4}+k2\pi\)
\(\Leftrightarrow x=-\dfrac{\pi}{48}+\dfrac{k}{2}\pi\)
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8.
\(cos7x-\sqrt{3}sin7x=-\sqrt{2}\)
\(\Leftrightarrow2\left(cos7x.\dfrac{1}{2}-sin7x.\dfrac{\sqrt{3}}{2}\right)=-\sqrt{2}\\ \Leftrightarrow cos\left(7x+\dfrac{\pi}{3}\right)=\dfrac{-\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2n\\7x+\dfrac{\pi}{3}=\dfrac{-3\pi}{4}+k2n\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{84}+\dfrac{k2\pi}{7}\\x=\dfrac{-13\pi}{84}+\dfrac{k2\pi}{7}\end{matrix}\right.\)
