1) $\cos{x} + \sqrt{3} \sin{x} = \sqrt{2} \Leftrightarrow \dfrac{1}{2} \cos{x} + \dfrac{\sqrt{3}}{2} \sin{x} = \dfrac{\sqrt{2}}{2} \\
\Leftrightarrow \cos{\left ( x - \dfrac{\pi}{6} \right )} = \cos{\left ( \dfrac{\pi}{4} \right )} \\
\Leftrightarrow
\left[\begin{matrix}
x - \dfrac{\pi}{6} = \dfrac{\pi}{4} + k2 \pi \\
x - \dfrac{\pi}{6} = - \dfrac{\pi}{4} + k2 \pi
\end{matrix}\right. \\
\Leftrightarrow .......$
2) $\sin{x} + \cos{x} = \dfrac{\sqrt{6}}{2} \Leftrightarrow \dfrac{1}{\sqrt{2}} \sin{x} + \dfrac{1}{\sqrt{2}} \cos{x} = \dfrac{\sqrt{3}}{2} \\
\Leftrightarrow \cos{\left ( x - \dfrac{\pi}{4} \right )} = \cos{\left ( \dfrac{\pi}{6} \right )} \\
\Leftrightarrow
\left[\begin{matrix}
x - \dfrac{\pi}{4} = \dfrac{\pi}{6} + k2 \pi \\
x - \dfrac{\pi}{4} = - \dfrac{\pi}{6} + k2 \pi
\end{matrix}\right. \\
\Leftrightarrow .......$
3)
C1: $\sin{x} + \cos{x} = \sqrt{2} \sin{5x} \Leftrightarrow \dfrac{1}{\sqrt{2}} \sin{x} + \dfrac{1}{\sqrt{2}} \cos{x} = \sin{5x} \\
\Leftrightarrow \sin{\left ( x + \dfrac{\pi}{4} \right )} = \sin{5x} \\
\Leftrightarrow
\left[\begin{matrix}
x + \dfrac{\pi}{4} = 5x + k2 \pi \\
x + \dfrac{\pi}{4} = \pi - 5x + k2 \pi
\end{matrix}\right. \\
\Leftrightarrow .......$
C2: $\sin{x} + \cos{x} = \sqrt{2} \sin{5x} \Leftrightarrow \dfrac{1}{\sqrt{2}} \sin{x} + \dfrac{1}{\sqrt{2}} \cos{x} = \sin{5x} \\
\Leftrightarrow \dfrac{1}{\sqrt{2}} \sin{x} + \dfrac{1}{\sqrt{2}} \cos{x} = \cos{\left ( \dfrac{\pi}{2} - 5x \right )} \\
\Leftrightarrow \cos{\left ( x - \dfrac{\pi}{4} \right )} = \cos{\left ( \dfrac{\pi}{2} - 5x \right )} \\
\Leftrightarrow
\left[\begin{matrix}
x - \dfrac{\pi}{4} = \dfrac{\pi}{2} - 5x + k2 \pi \\
x - \dfrac{\pi}{4} = - \left ( \dfrac{\pi}{2} - 5x \right ) + k2 \pi
\end{matrix}\right. \\
\Leftrightarrow
\left[\begin{matrix}
x - \dfrac{\pi}{4} = \dfrac{\pi}{2} - 5x + k2 \pi \\
x - \dfrac{\pi}{4} = - \dfrac{\pi}{2} + 5x + k2 \pi
\end{matrix}\right. \\
\Leftrightarrow .......$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
