giải PTLG ze ot!!

T

truongduong9083

sin4x=12sin4x = \dfrac{1}{2}
[4x=π6+k2π4x=5π6+k2π\Leftrightarrow \left[ \begin{array}{l} 4x= \dfrac{\pi}{6}+k2\pi \\ 4x= \dfrac{5\pi}{6}+k2\pi \end{array} \right.
[x=π24+kπ2x=5π24+kπ2\Leftrightarrow \left[ \begin{array}{l} x= \dfrac{\pi}{24}+ k\dfrac{\pi}{2} \\ x= \dfrac{5\pi}{24}+ k\dfrac{\pi}{2}\end{array} \right.
 
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