1 câu trả lời
$\tan\left({2x+\dfrac{\pi}{3}}\right)+2=5$
$\Rightarrow \tan\left({2x+\dfrac{\pi}{3}}\right)=3$
Đk: \(\cos\left({2x+\dfrac{\pi}{3}}\right)\ne0\)
\(\Rightarrow 2x+\dfrac{\pi}{3}\ne\dfrac{\pi}{2}+k\pi\)
$\Rightarrow x\ne\dfrac{\pi}{12}+k\dfrac{\pi}{2}$ $(k\in\mathbb Z)$
Pt $\Rightarrow 2x+\dfrac{\pi}{3}=\arctan3+k\pi$
$\Rightarrow x=-\dfrac{\pi}{6}+\dfrac{1}{2}.\arctan3+k\dfrac{\pi}{2},(k\in\mathbb Z)$.
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