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$\sin(2x-\dfrac{\pi}{4})=-3x-\dfrac{\pi}{3}$ $\Rightarrow \left[ \begin{array}{l} 2x-\dfrac{\pi}{4}=-3x-\dfrac{\pi}{3}+k2\pi \\ 2x-\dfrac{\pi}{4}=\pi+3x+\dfrac{\pi}{3}+k2\pi \end{array} \right .$ $\Rightarrow \left[ \begin{array}{l} x=-\dfrac{\pi}{60}+k\dfrac{2\pi}{5} \\ x=\dfrac{-19\pi}{12}+k2\pi \end{array} \right .(k\in\mathbb Z)$
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