1,Cos (3x+1)= sin $\frac{x}{2}$ 2, Cos ( $\frac{x}{2}$ – pi/3) = $\frac{1}{4}$ 01/10/2021 Bởi Kinsley 1,Cos (3x+1)= sin $\frac{x}{2}$ 2, Cos ( $\frac{x}{2}$ – pi/3) = $\frac{1}{4}$
1) \(\cos(3x+1)=\sin\dfrac{x}{2}\) \(\Leftrightarrow \cos(3x+1)=\cos\left({\dfrac{\pi}{2}-\dfrac{x}{2}}\right)\) \(\Leftrightarrow \left[ \begin{array}{l} 3x+1= \dfrac{\pi}{2}-\dfrac{x}{2}+k2\pi\\ 3x+1= -\left({\dfrac{\pi}{2}-\dfrac{x}{2}}\right)+k2\pi \end{array} \right .\) \(\Leftrightarrow \left[ \begin{array}{l} x=-1+\dfrac{\pi}{7}+k\dfrac{4\pi}{7} \\ x=-1-\dfrac{\pi}{5}+k\dfrac{4\pi}{5} \end{array} \right .\) \((k\mathbb Z)\) 2) \(\cos\left({\dfrac{x}{2}-\dfrac{\pi}{3}}\right)=\dfrac{1}{4}\) \(\Leftrightarrow \dfrac{x}{2}-\dfrac{\pi}{3}=\pm\arccos\dfrac{1}{4}+k2\pi\) \((k\in\mathbb Z)\). Bình luận
1) \(\cos(3x+1)=\sin\dfrac{x}{2}\)
\(\Leftrightarrow \cos(3x+1)=\cos\left({\dfrac{\pi}{2}-\dfrac{x}{2}}\right)\)
\(\Leftrightarrow \left[ \begin{array}{l} 3x+1= \dfrac{\pi}{2}-\dfrac{x}{2}+k2\pi\\ 3x+1= -\left({\dfrac{\pi}{2}-\dfrac{x}{2}}\right)+k2\pi \end{array} \right .\)
\(\Leftrightarrow \left[ \begin{array}{l} x=-1+\dfrac{\pi}{7}+k\dfrac{4\pi}{7} \\ x=-1-\dfrac{\pi}{5}+k\dfrac{4\pi}{5} \end{array} \right .\) \((k\mathbb Z)\)
2) \(\cos\left({\dfrac{x}{2}-\dfrac{\pi}{3}}\right)=\dfrac{1}{4}\)
\(\Leftrightarrow \dfrac{x}{2}-\dfrac{\pi}{3}=\pm\arccos\dfrac{1}{4}+k2\pi\) \((k\in\mathbb Z)\).