1.
\(\Leftrightarrow\dfrac{4}{5}sinx-\dfrac{3}{5}cosx=\dfrac{2}{5}\)
Đặt \(\dfrac{4}{5}=cosa\) với \(0< a< \dfrac{\pi}{2}\)
\(\Rightarrow sinx.cosa-cosx.sina=\dfrac{2}{5}\)
\(\Leftrightarrow sin\left(x-a\right)=\dfrac{2}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-a=arcsin\left(\dfrac{2}{5}\right)+k2\pi\\x-a=\pi-arcsin\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=a+arcsin\left(\dfrac{2}{5}\right)+k2\pi\\x=a+\pi-arcsin\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)
3.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin3x+\dfrac{1}{2}cos3x=\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(3x+\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{3}\\x=\dfrac{2\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
5.
Câu này chắc đề in nhầm, phải có 1 cái là sin2x thì hợp lý hơn.
\(\Leftrightarrow-7cos2x=13\)
\(\Leftrightarrow cos2x=-\dfrac{13}{7}< -1\)
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