a/
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=\sqrt{3}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\sqrt{\frac{3}{2}}>1\)
Pt vô nghiệm
b/
\(\Leftrightarrow\frac{2}{\sqrt{13}}sinx+\frac{3}{\sqrt{13}}cosx=\frac{2}{\sqrt{13}}\)
Đặt \(\frac{2}{\sqrt{13}}=cosa\) với \(a\in\left(0;\pi\right)\)
\(\Rightarrow sinx.cosa+cosx.sina=cosa\)
\(\Leftrightarrow sin\left(x+a\right)=sin\left(\frac{\pi}{2}-a\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+a=\frac{\pi}{2}-a+k2\pi\\x+a=\frac{\pi}{2}+a+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}-2a+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x=\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=x+\frac{\pi}{6}+k2\pi\\2x-\frac{\pi}{3}=\frac{5\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{7\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
2.
Theo điều kiện có nghiệm của pt lượng giác bậc nhất với sin và cos:
\(m^2+\left(m-1\right)^2\ge5\)
\(\Leftrightarrow m^2-m-2\ge0\Leftrightarrow\left[{}\begin{matrix}m\ge2\\m\le-1\end{matrix}\right.\)
