1.
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx+\frac{\sqrt{2}}{2}=0\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)+\frac{\sqrt{2}}{2}=0\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=-\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k2\pi\\x=\frac{17\pi}{12}+k2\pi\end{matrix}\right.\)
2.
\(\Leftrightarrow\frac{3}{\sqrt{13}}sin2x+\frac{2}{\sqrt{13}}cos2x=\frac{3}{\sqrt{13}}\)
Đặt \(\frac{3}{\sqrt{13}}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow sin2x.cosa+cos2x.sina=cosa\)
\(\Leftrightarrow sin\left(2x+a\right)=sin\left(\frac{\pi}{2}-a\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+a=\frac{\pi}{2}-a+k2\pi\\2x+a=\frac{\pi}{2}+a+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}-a+k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
3.
\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}\)
\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{3}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7\pi}{12}+k2\pi\\x=\frac{13\pi}{12}+k2\pi\end{matrix}\right.\)
4.
Câu này giống hệt câu a