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ĐK: $\sin(2x+\dfrac{1}{6})\ne 0\to x\ne \dfrac{-1}{12}+\dfrac{k\pi}{2}$
$\cot(2x+\dfrac{1}{6})=\tan\dfrac{1}{3}=\cot(\dfrac{\pi}{2}-\dfrac{1}{3})$
$\to 2x+\dfrac{1}{6}=\dfrac{\pi}{2}-\dfrac{1}{3}+k\pi$
$\to x=\dfrac{\pi}{4}-\dfrac{1}{4}+\dfrac{k\pi}{2}$ (TMĐK)
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